If the sum of an infinite GP a, a r, a r^2, a r^3, is 15 and the sum of the squares of its each term is 150 , then the sum of a r^2, a r^4, a r^6, is :

Correct answer: B. (1)/(2) Sum of infinite terms of series a +a r+a r^2+=15 (a)/(1-r)=15 ...(i) Sequence formed by square of terms: a^2, a^2 r^2, a^2 r^4, a^2 r^6 a^2+a^2 r^2+ =150 Sum =(a^2)/(1-r^2)=150 (a)/(1-r) · (a)/(1+r)=15 15 · (a)/(1+r)=150 ( a 1-r = 15) ⇒ (a)/(1+r)= 10 ...(ii) On dividing equation (i) by (ii) 1+r 1-r =(15)/(10) or r=(1)/(5) = 12 Now series : a r²+a r⁴+a r⁶+ Sum =(a^2)/(1-r^2)=(12 ·((1)/(25)))/(1-(1)/(25))=(1)/(2)

BITSAT 2023Paper 1

Exam: BITSAT
Year: 2023
Paper: Paper 1
Subject: Mathematics
Chapter: Sequences and Series

BITSAT 2023 · Paper 1 · Mathematics · Sequences and Series

If the sum of an infinite GP a, a r, a r^2, a r^3, is 15 and the...

If the sum of an infinite GP

a, a r, a r^2, a r^3, \ldots

is 15 and the sum of the squares of its each term is 150 , then the sum of

a r^2, a r^4, a r^6, \ldots

is :

Options

A $\frac{5}{2}$
B $\frac{1}{2}$
C $\frac{25}{2}$
D $\frac{9}{2}$

Correct answer

B. $\frac{1}{2}$

Step-by-step solution

Sum of infinite terms of series a

+a r+a r^2+=15

\therefore \frac{a}{1-r}=15

...(i)
Sequence formed by square of terms:

a^2, a^2 r^2, a^2 r^4, a^2 r^6 \ldots

a^2+a^2 r^2+\cdots=150

Sum

=\frac{a^2}{1-r^2}=150

\Rightarrow \frac{a}{1-r} \cdot \frac{a}{1+r}=15 \Rightarrow 15 \cdot \frac{a}{1+r}=150 \left(\because \frac{\mathrm{a}}{1-\mathrm{r}}=

15\right)

<br /> ⇒

\frac{a}{1+r}=

...(ii)<br />On dividing equation (i) by (ii)<br />

\frac{1+\mathrm{r}}{1-\mathrm{r}}=\frac{15}{10}

<br />or

r=\frac{1}{5} \Rightarrowa=

<br />Now series :

a r²+a r⁴+a r⁶+\ldots

<br />Sum

=\frac{a^2}{1-r^2}=\frac{12 \cdot\left(\frac{1}{25}\right)}{1-\frac{1}{25}}=\frac{1}{2}

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