JEE Mains Maths · Sequences and Series

Geometric Progressions: Terms and Sums

A geometric progression multiplies by the same ratio r at every step: find terms and sums from two conditions, use products and odd–even parts, and spot a GP hidden in a function, a recurrence or a two-base sum.

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Why this matters

Thirty-six PYQs. Most give two conditions on an increasing GP and ask for a later term or a sum. The rest hide the GP: in a function with f(x + y) = f(x)f(y), in a recurrence, or in a sum of products of powers of two bases. Three ideas cover the page.

Concept 1 of 3: Terms and sums of a GP from two conditions

a_n=ar^{n-1}

and

S_n=a\frac{r^n-1}{r-1}

. Two conditions give two equations in

a

and

r

: divide one by the other and

a

cancels, leaving

r

. A product condition is often quicker: terms whose positions add to the same total have the same product,

a_ma_n=a_pa_q

when

m+n=p+q

, so a product such as

a_3a_5

a_4^2

Definition

$a_n=ar^{n-1}$ , $S_n=a\frac{r^n-1}{r-1}$ ( $r\ne1$ ).
Divide two conditions to cancel $a$ .
$a_ma_n=a_pa_q$ when $m+n=p+q$ ; in particular $a_{k-j}a_{k+j}=a_k^2$ .
Three terms in GP: take $\frac ar,a,ar$ , so their product is $a^3$ .

Sum of n terms

S_n=a\,\frac{r^n-1}{r-1}

Worked example

A GP of positive terms has

a_3=12

and

a_5=48

. Find

S_6

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q57Moderate

Example 1 · Sequences and Series · GP: Terms and Sums

Let

a_{1},a_{2},a_{3}\ldots

. be a G.P. of increasing positive terms. If

a_{1}a_{5}= 28

and

a_{2}+a_{4}= 29

, the

a_{6}

is equal to

Increasing and positive means r > 1

Dividing two conditions often gives a quadratic in

r

with roots

t

and

\frac1t

. An increasing GP of positive terms keeps only the root above 1.

Drill 20 more on terms and sums of a gp from two conditions

Concept 2 of 3: Products, means and odd–even parts of a GP

The product of the first

n

terms is

a^nr^{n(n-1)/2}

, so its

n

th root is

ar^{(n-1)/2}

: the geometric mean of the terms is the middle term. The odd-placed terms form a GP with ratio

r^2

, and so do the even-placed ones, each even term being

r

times the odd term before it. The logarithms of GP terms are in AP.

Definition

Product of $n$ terms: $a^nr^{n(n-1)/2}$ .
Geometric mean of the terms $=ar^{(n-1)/2}$ .
$2m$ terms: (even-placed sum) $=r\times$ (odd-placed sum), so the whole sum is $(1+r)$ times the odd part.
$\log a_1,\log a_2,\dots$ are in AP with difference $\log r$ .

Whole sum of 2m terms

S=(1+r)\,S_{\text{odd}}

Worked example

A GP of 10 terms has sum 4 times the sum of its odd-placed terms. Find

r

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q61Moderate

Example 2 · Sequences and Series · GP: Terms and Sums

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

An odd number of terms

With an odd count there is one more odd-placed term than even-placed, and the factor

1+r

no longer holds. Write both sums out.

Drill 3 more on products, means and odd–even parts of a gp

Concept 3 of 3: Recognising a GP in a function, a recurrence or a two-base sum

Several shapes hide a GP. A function with

f(x+y)=f(x)f(y)

on natural numbers has

f(n)=f(1)^n

. A sum

\sum 2^k3^{m-k}

is a GP with ratio

\frac23

. A recurrence

a_{n+2}=pa_{n+1}+qa_n

is solved by powers of the roots of

t^2=pt+q

; conversely

a_n=\alpha^n-\beta^n

satisfies that recurrence. To find the remainder of a GP's sum, add the GP first, then reduce.

Definition

$f(x+y)=f(x)f(y)\Rightarrow f(n)=f(1)^n$ , a GP.
$\sum_{k=0}^{m}x^ky^{m-k}=\frac{y^{m+1}-x^{m+1}}{y-x}$ .
$a_n=A\alpha^n+B\beta^n$ where $\alpha,\beta$ are the roots of $t^2=pt+q$ .
Remainders: sum the GP in closed form, then reduce the closed form.

Two-base sum

\sum_{k=0}^{m}x^{k}y^{m-k}=\frac{y^{m+1}-x^{m+1}}{y-x}

Worked example

f(x+y)=f(x)f(y)

for natural

x,y

and

f(1)=2

. Find

\sum_{k=1}^{8}f(k)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q163Moderate

Example 3 · Sequences and Series · GP: Terms and Sums

Let

f(x)

be a function such that

f(x + y) = f(x) \cdot f(y)

for all

x,y \in N

. If

f(1) = 3

and

\sum_{k = 1}^{n} f(k) = 3279

, then the value of

n

f(x + y) = 2f(x)f(y) is not f(1)^n

Put

g=2f

: then

g(x+y)=g(x)g(y)

, so

g(n)=g(1)^n

and

f(n)=\frac12(2f(1))^n

Drill 10 more on recognising a gp in a function, a recurrence or a two-base sum

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Terms and sums of a GP from two conditions
Sum of n terms
$S_n=a\,\frac{r^n-1}{r-1}$
Products, means and odd–even parts of a GP
Whole sum of 2m terms
$S=(1+r)\,S_{\text{odd}}$
Recognising a GP in a function, a recurrence or a two-base sum
Two-base sum
$\sum_{k=0}^{m}x^{k}y^{m-k}=\frac{y^{m+1}-x^{m+1}}{y-x}$

Watch out for (3)

Increasing and positive means r > 1
→ Terms and sums of a GP from two conditions
An odd number of terms
→ Products, means and odd–even parts of a GP
f(x + y) = 2f(x)f(y) is not f(1)^n
→ Recognising a GP in a function, a recurrence or a two-base sum

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