Question

a=2 sum of first 4 terms is equal to one fourth of the sum of next 5 terms find the sym of first 30 terms​

Answer 1

Step-by-step explanation:

Given:

First term a = 2

Sum of next 5 terms will be equal to the difference between sum of 9 terms and sum of four terms.

Thus according to the given condition, we have:

$S_4 = \frac{1}{4} (S_9 - S_4) \\ S_4 = \frac{1}{4} [ \frac{9}{2} \{2 \times 2 + (9 - 1)d \} - \frac{4}{2} \{2 \times 2 + (4 - 1)d \}] \\ = \frac{1}{4} [ \frac{9}{2} \{4 + 8d \} - 2 \{4 +3d \}] \\ = \frac{1}{4} [ 18 + 36d - 8 - 6d ] \\ S_4 = \frac{1}{4} [10 + 30d] \\ \therefore \: \frac{4}{2} \{2 \times 2 + (4 - 1)d \}= \frac{1}{4} [10 + 30d] \\ 2 \{4 + 3d \}= \frac{1}{4} [10 + 30d] \\ 2 \times 4 \{4 + 3d \}= 10 + 30d \\ 8 \{4 + 3d \}= 10 + 30d \\ 32 + 24d = 10 + 30d \\ 32 - 10 = 30d - 24d \\ 22 = 6d \: \: \implies \: d = \frac{22}{6} \\ \implies \: d = \frac{11}{3} \\ hence \\ S_{30}= \frac{30}{2} \{2 \times 2 + (30 - 1) \times \frac{11}{3} \} \\ S_{30}= 15\{4+ 29 \times \frac{11}{3} \}\\ S_{30}= 15\{4+ \frac{319}{3} \} \\ S_{30}= 15\{\frac{3 \times 4 + 319}{3} \} \\ S_{30}= 15\{\frac{12 + 319}{3} \} \\ S_{30}= 15\{\frac{331}{3} \} \\ S_{30}= 5 \times 331 \\ S_{30}= 1655$