Sequences and Series 2 Question 1

1. If a1,a2,a3,… are in AP such that a1+a7+a16=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

Show Answer

Answer:

Correct Answer: 1. (a)

Solution:

  1. Let the common difference of given AP is ’ d ‘. Since,

a1+a7+a16=40

∴a1+a1+6d+a1+15d=40[∵an=a1+(n−1)d]

⇒3a1+21d=40 Now, sum of first 15 terms is given by

S15=152[2a1+(15−1)d]=152[2a1+14d]=15[a1+7d]

From Eq. (i), we have

a1+7d=403 So, S15=15×403=5×40=200

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