Let a, a r, a r², ............ be an infinite G.P. If Σn=0^∞ a rⁿ=57 and Σn=0^∞ a³ r³ⁿ=9747, then a+18 r is equal to

Solution

Maths sir · 4 steps

Cube the first sum and divide by the second: a cancels and you get an equation in r alone.

1. Write both sums. For |r| < 1, the first series gives a/(1-r) = 57. The second is a G.P. with first term a³ and ratio r³, so a³/(1-r³) = 9747.
2. Remove a. Cube the first result: a³/(1-r)³ = 57³ = 185193. Divide this by the second result, and a³ cancels. Since 185193 = 19 × 9747, the ratio is 19.

   (1-r³)/(1-r)³ = (1+r+r²)/(1-r)² = 19

3. Solve for r. Cross-multiply: 1 + r + r² = 19 - 38r + 19r², so 18r² - 39r + 18 = 0, that is 6r² - 13r + 6 = 0. The roots are r = 3/2 and r = 2/3. The series converges only for |r| < 1, so r = 2/3.
4. So. a = 57(1 - 2/3) = 19, and a + 18r = 19 + 12 = 31, which is option (c).

The idea to keep. When two infinite G.P. sums use powers of the same terms, raise one sum to the matching power and divide so that a cancels. Then use |r| < 1 to choose the right root.

Answer from the NTA final answer key. Solution by PrepWiser.