Sequences and Series 2 Question 1
1. If $a _1, a _2, a _3, \ldots$ are in $AP$ such that $a _1+a _7+a _{16}=40$, then the sum of the first 15 terms of this AP is
(2019 Main, 12 April II)
(a) 200
(b) 280
(c) 120
(d) 150
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Answer:
Correct Answer: 1. (a)
Solution:
- Let the common difference of given $AP$ is ’ $d$ ‘. Since,
$$ a _1+a _7+a _{16}=40 $$
$\therefore a _1+a _1+6 d+a _1+15 d=40 \quad\left[\because a _n=a _1+(n-1) d\right]$
$\Rightarrow 3 a _1+21 d=40$ Now, sum of first 15 terms is given by
$$ \begin{aligned} S _{15} & =\frac{15}{2}\left[2 a _1+(15-1) d\right] \\ & =\frac{15}{2}\left[2 a _1+14 d\right]=15\left[a _1+7 d\right] \end{aligned} $$
From Eq. (i), we have
$$ \begin{aligned} a _1+ & 7 d=\frac{40}{3} \\ \text { So, } \quad S _{15} & =15 \times \frac{40}{3} \\ & =5 \times 40=200 \end{aligned} $$
