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Sequences and Series question

2021 · 22 Jul · Shift 2 · Q24
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  5. /2021 · 22 Jul · Shift 2 · Q24

Sequences and Series question

2021 · 22 Jul · Shift 2 · Q24

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
Let Sn denote the sum of first n-terms of an arithmetic progression. If S10 = 530, S5 = 140, then S20 −-− S6 is equal to:
  1. A
    1862
  2. B
    1842
  3. C
    1852
  4. D
    1872
View written solutionFree

Correct answer: A

  1. Let the arithmetic progression have first term aaa and common difference ddd.

  2. Sum of first nnn terms of an AP is Sn=n2[2a+(n−1)d].S_n = \frac{n}{2}\left[2a + (n-1)d\right].Sn​=2n​[2a+(n−1)d].

  3. Use the given values:

    For n=10n=10n=10, S10=102[2a+9d]=5(2a+9d)=530S_{10} = \frac{10}{2}[2a+9d] = 5(2a+9d)=530S10​=210​[2a+9d]=5(2a+9d)=530 2a+9d=106.(1)2a+9d=106. \quad (1)2a+9d=106.(1)

    For n=5n=5n=5, S5=52[2a+4d]=140S_5 = \frac{5}{2}[2a+4d]=140S5​=25​[2a+4d]=140 2a+4d=56.(2)2a+4d=56. \quad (2)2a+4d=56.(2)

  4. Subtract (2) from (1): (2a+9d)−(2a+4d)=106−56 (2a+9d)-(2a+4d)=106-56(2a+9d)−(2a+4d)=106−56 5d=50⇒d=10.5d=50 \Rightarrow d=10.5d=50⇒d=10.

  5. Substitute d=10d=10d=10 into (2): 2a+40=562a+40=562a+40=56 2a=16⇒a=8.2a=16 \Rightarrow a=8.2a=16⇒a=8.

  6. Now find S20S_{20}S20​: S20=202[2a+19d]S_{20}=\frac{20}{2}[2a+19d]S20​=220​[2a+19d] =10[2(8)+19(10)]=10[2(8)+19(10)]=10[2(8)+19(10)] =10[16+190]=10⋅206=2060.=10[16+190]=10\cdot 206=2060.=10[16+190]=10⋅206=2060.

  7. Find S6S_6S6​: S6=62[2a+5d]S_6=\frac{6}{2}[2a+5d]S6​=26​[2a+5d] =3[16+50]=3⋅66=198.=3[16+50]=3\cdot 66=198.=3[16+50]=3⋅66=198.

  8. Therefore, S20−S6=2060−198=1862.S_{20}-S_6=2060-198=1862.S20​−S6​=2060−198=1862.

  9. Checking options: 186218621862 corresponds to Option A.

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