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Sequences and Series question

2019 · 12 Apr · Shift 1 · Q34
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Sequences and Series question

2019 · 12 Apr · Shift 1 · Q34

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let Sn denote the sum of the first n terms of an A.P. If S4 = 16 and S6= – 48, then S10 is equal to :
  1. A
    - 320
  2. B
    - 380
  3. C
    - 460
  4. D
    - 210
View written solutionFree

Correct answer: A

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

  2. Sum of first nnn terms of an A.P. is

Sn=n2[2a+(n−1)d].S_n=\frac{n}{2}\left[2a+(n-1)d\right].Sn​=2n​[2a+(n−1)d].
  1. Use the given values.

For n=4n=4n=4:

S4=42[2a+3d]=2(2a+3d)=16S_4=\frac{4}{2}[2a+3d]=2(2a+3d)=16S4​=24​[2a+3d]=2(2a+3d)=16 4a+6d=16  ⟹  2a+3d=8...(1)4a+6d=16 \implies 2a+3d=8 \quad ...(1)4a+6d=16⟹2a+3d=8...(1)

For n=6n=6n=6:

S6=62[2a+5d]=3(2a+5d)=−48S_6=\frac{6}{2}[2a+5d]=3(2a+5d)=-48S6​=26​[2a+5d]=3(2a+5d)=−48 6a+15d=−48...(2)6a+15d=-48 \quad ...(2)6a+15d=−48...(2)
  1. Solve equations (1) and (2).

From (1):

4a+6d=164a+6d=164a+6d=16

Multiply by 32\frac{3}{2}23​:

6a+9d=246a+9d=246a+9d=24

Now subtract from (2):

(6a+15d)−(6a+9d)=−48−24(6a+15d)-(6a+9d)=-48-24(6a+15d)−(6a+9d)=−48−24 6d=−72  ⟹  d=−126d=-72 \implies d=-126d=−72⟹d=−12

Substitute into (1):

2a+3(−12)=82a+3(-12)=82a+3(−12)=8 2a−36=82a-36=82a−36=8 2a=44  ⟹  a=222a=44 \implies a=222a=44⟹a=22
  1. Find S10S_{10}S10​:
S10=102[2a+9d]=5[2(22)+9(−12)]S_{10}=\frac{10}{2}[2a+9d]=5[2(22)+9(-12)]S10​=210​[2a+9d]=5[2(22)+9(−12)] =5[44−108]=5(−64)=−320=5[44-108]=5(-64)=-320=5[44−108]=5(−64)=−320
  1. Compare with options:
  • A: −320-320−320 ✅
  • B: −380-380−380
  • C: −460-460−460
  • D: −210-210−210

Therefore, the correct answer is A.

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