Easy Practice — Set 1

Quick Recap — Arithmetic Progressions

• $n$th term: $a_n=a+(n-1)d$; sum: $S_n=\tfrac n2\big(2a+(n-1)d\big)=\tfrac n2(a+l)$.
• Arithmetic mean of $a,b$ is $\tfrac{a+b}{2}$; $a,b,c$ in AP $\iff 2b=a+c$.
• Sum of first $n$ naturals $=\tfrac{n(n+1)}{2}$; sum of first $n$ odd numbers $=n^2$.
• Consecutive terms differ by the constant $d$.

Beyond-NCERT JEE Formulae

These five tools appear again and again in JEE Main but sit just past the NCERT core. Use them alongside the AP/GP basics you already know.

Harmonic Progression (HP) and Harmonic Mean

• A sequence is an HP exactly when the reciprocals of its terms are in AP. So if $t_1,t_2,t_3,\ldots$ is an HP, then $\dfrac{1}{t_1},\dfrac{1}{t_2},\dfrac{1}{t_3},\ldots$ is an AP.
• $n$th term: if that reciprocal AP has first term $a$ and common difference $d$, then $t_n=\dfrac{1}{a+(n-1)d}$. Never apply the AP or GP term formula directly to HP terms; always flip to reciprocals first.
• Harmonic mean of two positive numbers $a,b$ is $\mathrm{HM}(a,b)=\dfrac{2ab}{a+b}=\dfrac{2}{\tfrac{1}{a}+\tfrac{1}{b}}$. For $n$ numbers, $\mathrm{HM}=\dfrac{n}{\tfrac{1}{a_1}+\cdots+\tfrac{1}{a_n}}$.

AM $\ge$ GM $\ge$ HM

For positive numbers the three classical means satisfy $A\ge G\ge H$, with equality only when all the numbers are equal. For two numbers $a,b$: $A=\dfrac{a+b}{2}$, $G=\sqrt{ab}$, and $H=\dfrac{2ab}{a+b}$.

[JEE Tip] For two numbers the means are themselves linked by $$G^2=A\cdot H,$$ so $G$ is the geometric mean of $A$ and $H$. Given any two of $A,G,H$ this identity hands you the third at once, and it is a JEE favourite. (For three or more numbers only $A\ge G\ge H$ survives; the product identity is special to two numbers.)

Arithmetico-Geometric Progression (AGP)

An AGP multiplies an AP by a GP term-by-term: $t_n=[a+(n-1)d]\,r^{n-1}$.

• Finite sum (multiply and subtract): write out $S_n$, multiply every term by $r$ to get $rS_n$, and subtract. All the middle coefficients collapse into a plain GP, leaving one first term and one last term; then solve the linear equation for $S_n$.
• Infinite sum (converges for $|r|<1$): $$S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}.$$
• Two clean special cases worth memorising: $\displaystyle\sum_{n=1}^{\infty} n\,x^{n-1}=\dfrac{1}{(1-x)^2}$ and $\displaystyle\sum_{n=1}^{\infty} n^2 x^{n}=\dfrac{x(1+x)}{(1-x)^3}$, both for $|x|<1$.

Method of Differences (Telescoping)

If each term can be written as $f(k)-f(k+1)$, the whole sum collapses to its ends: $$\sum_{k=1}^{n}\big(f(k)-f(k+1)\big)=f(1)-f(n+1).$$

• $\displaystyle\sum_{k=1}^{n}\dfrac{1}{k(k+1)}=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}$, from $\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$.
• $\displaystyle\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{4}-\dfrac{1}{2(n+1)(n+2)}$, from $\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{2}\left(\dfrac{1}{k(k+1)}-\dfrac{1}{(k+1)(k+2)}\right)$.
• The same idea cracks $\dfrac{r}{(r+1)!}=\dfrac{1}{r!}-\dfrac{1}{(r+1)!}$ and denominators that factor as a difference of two quadratics.

Sum-Reciprocal Inequality

For any positive reals $a_1,a_2,\ldots,a_n$, $$(a_1+a_2+\cdots+a_n)\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots+\dfrac{1}{a_n}\right)\ge n^2,$$ with equality only when all the $a_i$ are equal. It is just $A\ge H$ rearranged (one line of Cauchy-Schwarz gives it too). The most-tested case is $n=3$: $(a+b+c)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge 9$.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Three means of two numbers

Problem. For the numbers $2$ and $8$, compute the AM, GM and HM, then verify $A\ge G\ge H$ and $G^2=A\cdot H$.

Solution. $A=\dfrac{2+8}{2}=5$, $G=\sqrt{2\cdot 8}=\sqrt{16}=4$, and $H=\dfrac{2\cdot 2\cdot 8}{2+8}=\dfrac{32}{10}=\dfrac{16}{5}=3.2$.

Ordering: $5\ge 4\ge 3.2$, so $A\ge G\ge H$ holds. Product check: $A\cdot H=5\cdot\dfrac{16}{5}=16=4^2=G^2$.

Answer. $A=5,\ G=4,\ H=\dfrac{16}{5}$; both relations confirmed.

Example 2 — Recover the HM and the numbers

Problem. Two positive numbers have AM $=13$ and GM $=12$. Find their HM and the numbers themselves.

Solution. By $G^2=A\cdot H$, the harmonic mean is $H=\dfrac{G^2}{A}=\dfrac{144}{13}$. For the numbers, $a+b=2A=26$ and $ab=G^2=144$, so they are the roots of $x^2-26x+144=0$. The discriminant is $26^2-4\cdot 144=676-576=100$, giving $x=\dfrac{26\pm 10}{2}$, i.e. $x=18$ or $x=8$.

Answer. $H=\dfrac{144}{13}$; the two numbers are $8$ and $18$.

Example 3 — Sum the AGP $\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{2^{n}}$

Problem. Evaluate $S=\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+\cdots$

Solution (multiply and subtract). Multiply $S$ by the ratio $\tfrac{1}{2}$ and line the two rows up: $$S=\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\cdots,\qquad \tfrac{1}{2}S=\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\cdots$$ Subtracting term by term, $S-\tfrac{1}{2}S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots=1$, an infinite GP of ratio $\tfrac{1}{2}$. Hence $\tfrac{1}{2}S=1$, so $S=2$. As a check, $\displaystyle\sum_{n=1}^{\infty} n\,x^{n-1}=\dfrac{1}{(1-x)^2}$ with $x=\tfrac{1}{2}$ gives $\tfrac{1}{2}\cdot\dfrac{1}{(1/2)^2}=2$.

Answer. $S=2$.

Example 4 — AGP by the infinite-sum formula

Problem. Evaluate $\displaystyle\sum_{n=1}^{\infty}\dfrac{2n-1}{3^{\,n-1}}=1+\dfrac{3}{3}+\dfrac{5}{9}+\dfrac{7}{27}+\cdots$

Solution. This is an AGP with AP part $a=1$, $d=2$ and GP ratio $r=\dfrac{1}{3}$ (so $|r|<1$). Apply $S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}$ with $1-r=\dfrac{2}{3}$: $$\dfrac{a}{1-r}=\dfrac{1}{2/3}=\dfrac{3}{2},\qquad \dfrac{dr}{(1-r)^2}=\dfrac{2\cdot\tfrac{1}{3}}{(2/3)^2}=\dfrac{2/3}{4/9}=\dfrac{3}{2}.$$ Adding the two pieces, $S_\infty=\dfrac{3}{2}+\dfrac{3}{2}=3$.

Answer. $S_\infty=3$.

Example 5 — Telescoping a triple product

Problem. Find $\displaystyle\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)}$, and hence its value for $n=8$ and as $n\to\infty$.

Solution. Split with $\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{2}\left(\dfrac{1}{k(k+1)}-\dfrac{1}{(k+1)(k+2)}\right)$. Writing $g(k)=\dfrac{1}{k(k+1)}$, the sum telescopes: $$\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{2}\big(g(1)-g(n+1)\big)=\dfrac{1}{4}-\dfrac{1}{2(n+1)(n+2)}.$$ For $n=8$ this is $\dfrac{1}{4}-\dfrac{1}{2\cdot 9\cdot 10}=\dfrac{1}{4}-\dfrac{1}{180}=\dfrac{45-1}{180}=\dfrac{44}{180}=\dfrac{11}{45}$. As $n\to\infty$ the second term vanishes, leaving $\dfrac{1}{4}$.

Answer. $\dfrac{1}{4}-\dfrac{1}{2(n+1)(n+2)}$; for $n=8$ it equals $\dfrac{11}{45}$, and the infinite sum is $\dfrac{1}{4}$.

Example 6 — Minimum via the sum-reciprocal inequality

Problem. If $a,b,c>0$ and $a+b+c=1$, find the least value of $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$.

Solution. The inequality $(a+b+c)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge 9$ gives, since $a+b+c=1$, $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9.$$ Equality needs $a=b=c$, i.e. $a=b=c=\dfrac{1}{3}$, for which $3+3+3=9$.

Answer. The minimum is $9$, attained at $a=b=c=\dfrac{1}{3}$.

Example 7 — Insert four harmonic means

Problem. Insert four harmonic means between $1$ and $\dfrac{1}{6}$.

Solution. For an HP we work with the reciprocals, which must form an AP. The reciprocals of the two ends are $\dfrac{1}{1}=1$ and $\dfrac{1}{1/6}=6$. With four means inserted the AP has $6$ terms, so its common difference is $d=\dfrac{6-1}{4+1}=1$. The reciprocal AP is therefore $1,2,3,4,5,6$; taking reciprocals back gives the HP $1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5},\dfrac{1}{6}$.

Answer. The four harmonic means are $\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},\dfrac{1}{5}$.

Example 8 — A cyclic AM-GM minimum

Problem. For positive reals $a,b,c$, find the minimum of $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$.

Solution. The three positive terms have product $\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}=1$. Applying AM $\ge$ GM to these three terms, $$\dfrac{1}{3}\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge\left(\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}\right)^{1/3}=1,$$ so $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge 3$, with equality when $a=b=c$.

Answer. The minimum value is $3$.