Zaymiey

📐 Mathematics  ·  Sequences and Series  ·  JEE

If the sum of an infinite GP is 3 and sum of squares of its terms is 9/5, find the series.

Answer: 1, 2/3, 4/9,....

  • A 1, 2/3, 4/9,...
  • B 2, 4/3, 8/9,...
  • C 3/2, 1, 2/3,...
  • D 1, 1, 1,...

Correct answer: A. 1, 2/3, 4/9,...

Explanation: Let first term a, ratio r. Sum: a/(1-r)=3. Sum of squares: a²/(1-r²)=9/5. Dividing the second by the square of the first: (1-r)/(1+r) = (9/5)/9 = 1/5. So 5(1-r)=1+r, giving 4=6r, r=2/3. Then a=3(1-2/3)=1. Series: 1, 2/3, 4/9, ...

Each bar grows by the common difference d=3a1=2a2=5a3=8a4=11d=3

Bar heights for AP 2, 5, 8, 11 with the common difference d=3 marked between consecutive terms.

Concept context

Arithmetic progressions, geometric progressions, and sums

Read the full Sequences and Series notes →