Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Sequences and Series question

2020 · 3 Sep · Shift 1 · Q34
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Sequences and Series
  5. /2020 · 3 Sep · Shift 1 · Q34

Sequences and Series question

2020 · 3 Sep · Shift 1 · Q34

JEE MainMathematicsSequences and SeriesMCQ+4 / −1
If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is :
  1. A
    14{1 \over 4}41​
  2. B
    15{1 \over 5}51​
  3. C
    17{1 \over 7}71​
  4. D
    16{1 \over 6}61​
View written solutionFree

Correct answer: D

  1. Let the A.P. have first term a=3a=3a=3 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].

So, S25=252[2⋅3+24d]=252(6+24d)=75+300d.S_{25}=\frac{25}{2}\left[2\cdot 3+24d\right]=\frac{25}{2}(6+24d)=75+300d.S25​=225​[2⋅3+24d]=225​(6+24d)=75+300d.

  1. The sum of the next 151515 terms means the sum from 26th to 40th term: T26+T27+⋯+T40=S40−S25.T_{26}+T_{27}+\cdots+T_{40}=S_{40}-S_{25}.T26​+T27​+⋯+T40​=S40​−S25​.

Given: S25=S40−S25S_{25}=S_{40}-S_{25}S25​=S40​−S25​ which gives 2S25=S40.2S_{25}=S_{40}.2S25​=S40​.

  1. Now compute S40S_{40}S40​: S40=402[2⋅3+39d]=20(6+39d)=120+780d.S_{40}=\frac{40}{2}\left[2\cdot 3+39d\right]=20(6+39d)=120+780d.S40​=240​[2⋅3+39d]=20(6+39d)=120+780d.

  2. Use 2S25=S402S_{25}=S_{40}2S25​=S40​: 2(75+300d)=120+780d2(75+300d)=120+780d2(75+300d)=120+780d 150+600d=120+780d150+600d=120+780d150+600d=120+780d 30=180d30=180d30=180d d=30180=16.d=\frac{30}{180}=\frac{1}{6}.d=18030​=61​.

  3. Therefore, the common difference is 16.\boxed{\frac{1}{6}}.61​​.

  4. Checking options: this corresponds to Option D.

PreviousNext

More from Sequences and Series

  • The value of (0.16)log2.5​(31​+321​+....to∞) is equal to ​.2020 · Numerical
  • If m arithmetic means (A.Ms) and three geometric means (G.Ms) are inserted between 3 and 243 such that 4th A.M. is equal to 2nd G.M., then m is equal to ​ .2020 · Numerical
  • The minimum value of 2sinx + 2cosx is :2020 · MCQ
  • Let a1, a2, ..., an be a given A.P. whose common difference is an integer and Sn = a1 + a2 + .... + an. If a1 = 1, an = 300 and 15 ≤ n ≤ 50, then the ordered pair (Sn-4, an–4) is equal to:2020 · MCQ
  • If 32sin2α−1, 14 and 34−2sin2α are the first three terms of an A.P. for some α, then the sixth terms of this A.P. is:2020 · MCQ
  • If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is :2020 · MCQ
  • Let a , b, c , d and p be any non zero distinct real numbers such that (a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then :2020 · MCQ
  • The common difference of the A.P. b1, b2, … , bm is 2 more than the common difference of A.P. a1, a2, …, an. If a40 = –159, a100 = –399 and b100 = a70, then b1 is equal to :2020 · MCQ