Question

The ratio of the sum of n terms of the two A.P's. be $\frac{7n + 1}{4n + 27}$ and ratio of 11th term is l then value of 111 × l is-

• A. 138
• B. 128
• C. 122
• D. 148

Answer 1

Answer: (D)

148

Answer 2

Let first A.P. is a₁, a₁ + d₁, a₁ + 2d₁…….

a₁(first term), d₁ (common difference)

Second A.P. is a₂ , a₂ + d₂, a₂ + 2d₂……

a₂ (first term), d₂ (common difference)

Given is $\frac{n/2\lbrack 2a_{1} + (n - 1)d_{1}\rbrack}{n/2\lbrack 2a_{2} + (n - 1)d_{2}\rbrack}$= $\frac{7n + 1}{4n + 27}$

Ž $\frac{a_{1} + \left( \frac{n - 1}{2} \right)d_{1}}{a_{2} + \left( \frac{n - 1}{2} \right)d_{2}}$=$\frac{7n + 1}{4n + 27}$

Put $\frac{n - 1}{2}$= 10 or n = 21 to get

$\frac{a_{1} + 10d_{1}}{a_{2} + 10d_{2}}$= $\frac{7 \times 21 + 1}{4 \times 21 + 27}$= $\frac{148}{111}$