Questions and Answers Section

Example 1: Finding a Specific Term in AP

Question: Find the 15th term of the AP: 3, 7, 11, 15, …

Solution: Here, $a=3, d=4, n=15$.

Using $a_n = a + (n-1)d$

$a_{15} = 3 + (14)4 = 59$.

Example 2: Finding the Sum of an AP

Question: Find the sum of the first 20 terms of the AP 1, 4, 7, 10, …

Solution:

Here, $a=1, d=3, n=20$. Using $S_n = \frac{n}{2}[2a + (n-1)d]$: $S_{20} = \frac{20}{2}[2(1) + 19(3)] = 10[59] = 590$.

Example 3: Finding a Specific Term in GP

Question: Find the 6th term of the GP: 5, 10, 20, 40, …

Solution:

Here, $a=5, r=2, n=6$. Using $a_n = ar^{n-1}$: $a_6 = 5 \cdot 2^5 = 160$.

Example 4: Finding the Sum of a GP

Question: Find the sum of the first 8 terms of the GP 1, 3, 9, 27, …

Solution:

Here, $a=1, r=3, n=8$.

Using $S_n = \frac{a(r^n-1)}{r-1}$: $S_8 = \frac{1(3^8-1)}{3-1} = \frac{6560}{2} = 3280$.

Example 5: Finding a Term in HP

Question: The 5th term of an HP is 1/12 and the 11th term is 1/24. Find the 16th term.

Solution:

In the corresponding AP, $a_5=12$ and $a_{11}=24$.

So, $a+4d=12$ and $a+10d=24$.

Solving gives $d=2, a=4$. The 16th term of the AP is $a_{16} = 4+15(2)=34$.

The 16th term of the HP is 1/34.

Example 6: Sum of an Infinite GP

Question: Find the sum of the infinite series $16 + 8 + 4 + 2 + ...$

Solution:

This is an infinite GP with $a=16, r=1/2$.

Since $|r| < 1$, the sum is $S_{\infty} = \frac{a}{1-r} = \frac{16}{1 - 1/2} = 32$.

Example 7: Classic AP Property Problem

Question: If the p-th term of an AP is q and the q-th term is p, find the r-th term.

Solution:

Given $a + (p-1)d = q$ and $a + (q-1)d = p$.

Subtracting gives $(p-q)d = q-p$, so $d=-1$.

Substituting back gives $a=p+q-1$.

The r-th term is $a_r = a+(r-1)d = (p+q-1) + (r-1)(-1) = p+q-r$.

Example 8: Selection of Terms in AP

Question: The sum of three numbers in an AP is 21, and the sum of their squares is 155. Find the numbers.

Solution:

Let the numbers be $a-d, a, a+d$.

Sum = $3a = 21 \implies a=7$.

Sum of squares = $(7-d)^2+7^2+(7+d)^2 = 155$

$\implies$ $49-14d+d^2+49+49+14d+d^2=155 \implies 147+2d^2=155 \implies 2d^2=8 \implies d=\pm 2$.

The numbers are 5, 7, 9.

Example 9: Selection of Terms in GP

Question: The sum of three numbers in a GP is 13/12, and their product is -1. Find the numbers.

Solution:

Let the numbers be $a/r, a, ar$.

Product = $a^3 = -1 \implies a = -1$.

Sum = $-1/r - 1 - r = 13/12$.

This gives $12r^2+25r+12=0$, with solutions $r=-3/4, -4/3$.

The numbers are 4/3, -1, 3/4.

Example 10: AM-GM Inequality

Question: Find the minimum value of $9x^2 + \frac{1}{x^2}$ for $x>0$.

Solution:

Using AM-GM: $\frac{9x^2 + 1/x^2}{2} \ge \sqrt{9x^2 \cdot \frac{1}{x^2}} = 3$.

So, $9x^2 + \frac{1}{x^2} \ge 6$. The minimum value is 6.

Example 11: Using G² = AH

Question: The AM between two numbers is 15 and their HM is 9.6. Find their GM.

Solution:

Using the identity $G^2 = AH$, we have $G^2 = 15 \times 9.6 = 144$. Therefore, the GM is $G = \sqrt{144} = 12$.

Example 12: Summation using $\Sigma$ Formulas

Question: Find the sum of the first n terms of the series $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ...$

Solution:

The n-th term is $a_n = n(n+1) = n^2+n$.

The sum is $S_n = \sum (k^2+k) = \sum k^2 + \sum k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3}$.

Example 13: Sum of an Infinite AGP

Question: Find the sum of the infinite series $1 + \frac{4}{5} + \frac{7}{25} + \frac{10}{125} + ...$

Solution:

This is an AGP with $a=1, d=3, r=1/5$.

Using $S_{\infty} = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$: $S_{\infty} = \frac{1}{4/5} + \frac{3(1/5)}{(4/5)^2} = \frac{5}{4} + \frac{15}{16} = \frac{35}{16}$.

Example 14: Method of Difference (Telescoping Series)

Question: Find the sum of n terms of the series $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ...$

Solution:

The n-th term is $a_n = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. The sum telescopes to $S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1}$.

Example 15: Inserting Arithmetic Means

Question: Insert 4 arithmetic means between 5 and 25.

Solution:

We create an AP of 6 terms where $a=5$ and $a_6=25$. $a_6=a+5d \implies 25=5+5d \implies d=4$. The means are 9, 13, 17, 21.

Example 16: Inserting Geometric Means

Question: Insert 3 geometric means between 1 and 256.

Solution:

We create a GP of 5 terms where $a=1$ and $a_5=256$. $a_5 = ar^4 \implies 256 = 1 \cdot r^4 \implies r=4$. The means are 4, 16, 64.

Example 17: AP and GP Condition

Question: If the 2nd, 5th, and 9th terms of a non-constant AP are in GP, find the common ratio of the GP.

Solution:

The terms are $a+d, a+4d, a+8d$. Since they are in GP, $(a+4d)^2 = (a+d)(a+8d)$. Solving gives $a=8d$. The common ratio is $\frac{a+4d}{a+d} = \frac{12d}{9d} = 4/3$.

Example 18: Sum of Cubes

Question: Find the sum of the series $1^3 + 2^3 + ... + 10^3$.

Solution:

Using the formula $\sum k^3 = [\frac{n(n+1)}{2}]^2$ with n=10. Sum = $(\frac{10 \times 11}{2})^2 = 55^2 = 3025$.

Example 19: Sum in HP

Question: Three numbers are in HP. The sum of the numbers is 37, and the sum of their reciprocals is 1/4. Find the numbers.

Solution:

Let the reciprocals be in AP: $a-d, a, a+d$.

Sum of reciprocals = $3a = 1/4 \implies a=1/12$.

The middle term of the HP is 12. Let the HP terms be x, 12, y.

We have $x+12+y=37 \implies x+y=25$.

Also, $1/6=(x+y)/xy \implies 1/6=25/xy \implies xy=150$.

The numbers are 10 and 15. The HP is 10, 12, 15.

Example 20: AGP with Negative Ratio

Question: Find the sum of the infinite series $1 - 2x + 3x^2 - 4x^3 + ...$ for $|x|<1$.

Solution:

This is the standard expansion for $(1+x)^{-2}$.

Let $S = 1 - 2x + 3x^2 - ...$

Using the $S-rS$ method with $r=-x$, we find $(1+x)S = 1-x+x^2-... = \frac{1}{1+x}$.

So, $S = \frac{1}{(1+x)^2}$.

Example 21: Three Numbers in AP and GP

Question: If three numbers are in AP and their sum is 15. If 1, 4, 19 are added to them respectively, they are in GP. Find the numbers.

Solution:

Let the numbers be $a-d, a, a+d$.

Sum is $3a=15 \implies a=5$.

The numbers are $5-d, 5, 5+d$.

The new numbers are $6-d, 9, 24+d$.

These are in GP, so $9^2 = (6-d)(24+d)$. $81 = 144 - 18d - d^2$.

This gives $d^2+18d-63=0 \implies (d+21)(d-3)=0$.

So $d=3$ or $d=-21$. The numbers are 2, 5, 8 or 26, 5, -16.

Example 22: Logarithms and GP

Question: If $a,b,c$ are the p-th, q-th, and r-th terms of a GP, prove that $(q-r)\log a + (r-p)\log b + (p-q)\log c = 0$.

Solution:

Let the GP have first term A and common ratio R.

$a=AR^{p-1}, b=AR^{q-1}, c=AR^{r-1}$.

Taking log: $\log a = \log A + (p-1)\log R$, etc. Substitute these into the given expression.

The terms involving $\log A$ sum to $(q-r+r-p+p-q)\log A = 0$.

The terms involving $\log R$ sum to $((q-r)(p-1)+(r-p)(q-1)+(p-q)(r-1))\log R$.

This expression simplifies to 0. Hence, the result is proved.

Example 23: Sum of Squares of an AP

Question: If $S_n$ denotes the sum of n terms of an AP, prove that $S_{3n} = 3(S_{2n}-S_n)$.

Solution:

RHS = $3[\frac{2n}{2}(2a+(2n-1)d) - \frac{n}{2}(2a+(n-1)d)] = \frac{3n}{2}[2(2a+2nd-d)-(2a+nd-d)] = \frac{3n}{2}[2a+3nd-d]$. This is the formula for $S_{3n}$.

Example 24: Finding Sum of a Special Series

Question: Find the sum of the series $1 + (1+2) + (1+2+3) + ...$ up to n terms.

Solution:

The k-th term is the sum of the first k natural numbers, $T_k = \frac{k(k+1)}{2} = \frac{1}{2}(k^2+k)$.

We need to find $S_n = \sum T_k = \frac{1}{2}(\sum k^2 + \sum k)$.

Substituting the standard formulas and simplifying gives $S_n = \frac{n(n+1)(n+2)}{6}$.

Example 25: Minimum Value with 3 Variables

Question: If a, b, c are positive real numbers, find the minimum value of $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$.

Solution:

By AM-HM inequality, $\frac{a+b+c}{3} \ge \frac{3}{1/a+1/b+1/c}$.

Rearranging gives $(a+b+c)(1/a+1/b+1/c) \ge 9$.

The minimum value is 9.

Example 26: AP with a Twist

Question: The sum of n terms of two APs are in the ratio $5n+4:9n+6$. Find the ratio of their 18th terms.

Solution:

Given $\frac{S_n}{S'_n} = \frac{n/2(2a_1+(n-1)d_1)}{n/2(2a_2+(n-1)d_2)} = \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2} = \frac{5n+4}{9n+6}$.

We want the ratio of the 18th terms, which is $\frac{a_{18}}{a'_{18}} = \frac{a_1+17d_1}{a_2+17d_2}$.

To get this form, we substitute $n=2(18)-1=35$ into the sum ratio expression.

$\frac{2a_1+(35-1)d_1}{2a_2+(35-1)d_2} = \frac{2(a_1+17d_1)}{2(a_2+17d_2)} = \frac{5(35)+4}{9(35)+6} = \frac{179}{321}$.

Example 27: Infinite GP with a variable

Question: Find the sum of the infinite series $x+y, x^2+xy+y^2, x^3+x^2y+xy^2+y^3, ...$ where $|x|<1, |y|<1$.

Solution:

The n-th term is $\frac{x^{n+1}-y^{n+1}}{x-y}$.

The sum is $\frac{1}{x-y} \sum (x^{n+1}-y^{n+1}) = \frac{1}{x-y}[ (\sum x^{n+1}) - (\sum y^{n+1}) ]$.

Both are infinite GPs. The sum is $\frac{1}{x-y}[ \frac{x^2}{1-x} - \frac{y^2}{1-y} ]$.

Example 28: Method of Difference with Factorials

Question: Find the sum of the series $\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + ...$ up to n terms.

Solution:

The n-th term is $T_n = \frac{n}{(n+1)!} = \frac{(n+1)-1}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$.

This is a telescoping series.

$S_n = (\frac{1}{1!}-\frac{1}{2!}) + (\frac{1}{2!}-\frac{1}{3!}) + ... + (\frac{1}{n!}-\frac{1}{(n+1)!}) = 1 - \frac{1}{(n+1)!}$.

Example 29: Sum of terms in HP

Question: If the first two terms of an HP are 2/5 and 12/13, find the largest term.

Solution:

The first two terms of the corresponding AP are 5/2 and 13/12.

The common difference is $d=13/12 - 5/2 = (13-30)/12 = -17/12$.

The AP is decreasing, so its terms will become negative.

The reciprocals (the HP terms) will approach negative infinity.

Therefore, the largest term is the first term, 2/5. This is a trick question.

Example 30: AP, GP, HP Condition

Question: If a,b,c are in AP, b,c,d are in GP and c,d,e are in HP, show that a,c,e are in GP.

Solution:

$2b=a+c$, $c^2=bd$, $d=\frac{2ce}{c+e}$.

From the GP, $d=c^2/b$.

From the AP, $b=(a+c)/2$.

Substitute b into the expression for d: $d=\frac{c^2}{(a+c)/2} = \frac{2c^2}{a+c}$.

Now equate the two expressions for d: $\frac{2c^2}{a+c} = \frac{2ce}{c+e}$.

This simplifies to $c(c+e)=e(a+c) \implies c^2+ce=ae+ce \implies c^2=ae$.

This is the condition for a, c, e to be in GP.