Sequences and Series 1 Question 1
1. If $a _1, a _2, a _3, \ldots, a _n$ are in $AP$ and $a _1+a _4+a _7+\ldots+a _{16}$ $=114$, then $a _1+a _6+a _{11}+a _{16}$ is equal to
(a) 64
(b) 76
(c) 98
(d) 38
(2019 Main, 10 April I)
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Answer:
Correct Answer: 1. (b)
Solution:
- Key Idea Use $n$th term of an AP i.e. $a _n=a+(n-1) d$, simplify the given equation and use result.
Given $AP$ is $a _1, a _2, a _3, \ldots, a _n$
Let the above AP has common difference ’ $d$ ‘, then $a _1+a _4+a _7+\ldots+a _{16}$
$=a _1+\left(a _1+3 d\right)+\left(a _1+6 d\right)+\ldots+\left(a _1+15 d\right)$
$=6 a _1+(3+6+9+12+15) d$
$$ \begin{aligned} & \therefore 6 a _1+45 d=114 \\ & \Rightarrow \quad 2 a _1+15 d=38 \end{aligned} $$
(given)
Now, $a _1+a _6+a _{11}+a _{16}$
$=a _1+\left(a _1+5 d\right)+\left(a _1+10 d\right)+\left(a _1+15 d\right)$
$=4 a _1+30 d=2\left(2 a _1+15 d\right)$
$=2 \times 38=76$
[from Eq. (i)]
