Sequences and Series 2 Question 4
4. If the sum and product of the first three terms in an AP are 33 and 1155, respectively, then a value of its 11 th term is
(2019 Main, 9 April II)
(a) 25
(b) -36
(c) -25
(d) -35
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Answer:
Correct Answer: 4. (b)
Solution:
- Let first three terms of an $AP$ as $a-d, a, a+d$.
So, $\quad 3 a=33 \Rightarrow a=11$
[given sum of three terms $=33$ and product of terms $=1155]$
$$ \begin{array}{cc} \Rightarrow & (11-d) 11(11+d)=1155 \\ \Rightarrow & 11^{2}-d^{2}=105 \\ \Rightarrow & d^{2}=121-105=16 \\ \Rightarrow & d= \pm 4 \end{array} $$
So the first three terms of the AP are either 7, 11, 15 or $15,11,7$.
So, the 11 th term is either $7+(10 \times 4)=47$
or $15+(10 \times(-4))=-25$.
Key Idea Use the formula of sum of first $n$ terms of AP, i.e $S _n=\frac{n}{2}[2 a+(n-1) d]$
Given AP, is
$a _1, a _2, a _3, \ldots$ having sum of first $n$-terms
$$ =\frac{n}{2}\left[2 a _1+(n-1) d\right] $$
[where, $d$ is the common difference of AP]
$$ =50 n+\frac{n(n-7)}{2} A $$
(given)
$\Rightarrow \quad \frac{1}{2}\left[2 a _1+(n-1) d\right]=50+\frac{n-7}{2} A$
$\Rightarrow \quad \frac{1}{2}\left[2 a _1+n d-d\right]=50-\frac{7}{2} A+\frac{n}{2} A$
$\Rightarrow \quad a _1-\frac{d}{2}+\frac{n d}{2}=50-\frac{7}{2} A+\frac{n}{2} A$ On comparing corresponding term, we get
$$ \begin{aligned} & d=A \text { and } a _1-\frac{d}{2}=50-\frac{7}{2} A \\ & \Rightarrow \quad a _1-\frac{A}{2}=50-\frac{7}{2} A \quad[\because d=A] \\ & \Rightarrow \quad a _1=50-3 A \\ & \text { So } \quad a _{50}=a _1+49 d \\ & \begin{array}{ll} =(50-3 A)+49 A & {[\because d=A]} \\ =50+46 A & \end{array} \end{aligned} $$
Therefore, $\left(d, a _{50}\right)=(A, 50+46 A)$
