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Sequences and Series question

2019 · 12 Apr · Shift 2 · Q30
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Sequences and Series question

2019 · 12 Apr · Shift 2 · Q30

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
If a1, a2, a3, ..... are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is :
  1. A
    120
  2. B
    200
  3. C
    150
  4. D
    280
View written solutionFree

Correct answer: B

Let the A.P. have first term aaa and common difference ddd.

Then, an=a+(n−1)da_n = a + (n-1)dan​=a+(n−1)d

1. Write the given terms

We have: a1=aa_1 = aa1​=a a7=a+6da_7 = a+6da7​=a+6d a16=a+15da_{16} = a+15da16​=a+15d

Given: a1+a7+a16=40a_1 + a_7 + a_{16} = 40a1​+a7​+a16​=40

So, a+(a+6d)+(a+15d)=40a + (a+6d) + (a+15d) = 40a+(a+6d)+(a+15d)=40 3a+21d=403a + 21d = 403a+21d=40 a+7d=403a + 7d = \frac{40}{3}a+7d=340​

2. Find the sum of first 15 terms

Sum of first 151515 terms of an A.P. is S15=152[2a+14d]S_{15} = \frac{15}{2}[2a + 14d]S15​=215​[2a+14d] S15=15(a+7d)S_{15} = 15(a+7d)S15​=15(a+7d)

Using a+7d=403a+7d = \frac{40}{3}a+7d=340​, S15=15⋅403=5⋅40=200S_{15} = 15 \cdot \frac{40}{3} = 5 \cdot 40 = 200S15​=15⋅340​=5⋅40=200

3. Compare with options

Thus, S15=200S_{15} = 200S15​=200

So the correct option is B.

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