Question

if the sum of first 4 terms of an AP is 40 and that of first 14 terms is280, find the sum of first n terms

Answer 1

Answer:

n^2 + 6n

Step-by-step explanation:

$S_4 = 40 \\ \\ \frac{4}{2}[a_1 + a_4] = 40 \\ \\ 2[a_1 + a_4] = 40 \\ \\ a_1 + a_4 = 40 \div 2 = 20 \ \ \ \longrightarrow \ \ \ (1)$

$S_{14} = 280 \\ \\ \frac{14}{2}[a_1 + a_{14}] = 280 \\ \\ 7[a_1+ a_{14}] = 280 \\ \\ a_1 + a_{14} = 280 \div 7 = 40 \ \ \ \longrightarrow \ \ \ (2)$

$(2) - (1) \\ \\ (a_1 + a_{14}) - (a_1 + a_4) = 40 - 20 \\ \\ a_1 + a_{14} - a_1 - a_4 = 40 - 20 \\ \\ a_{14} - a_4 = 20 \\ \\ (14 - 4)d = 20 \\ \\ 10d = 20 \\ \\ d = 20 \div 10 = 2$

$From \ (1), \\ \\ a_1 + a_4 = 20 \\ \\ a_1 + a_1 + 3d = 20 \ \ \ \ \ [a_4 = a_1 + (4 - 1)d = a_1 + 3d] \\ \\ 2a_1 + 3d = 20 \\ \\ 2a_1 + 3 \times 2 = 20 \\ \\ 2a_1 + 6 = 20 \\ \\ 2a_1 = 20 - 6 = 14 \\ \\ a_1 = 14 \div 2 = 7$

$S_n \\ \\ \frac{n}{2}[2a_1 + (n - 1)d] \\ \\ \frac{n}{2}[2 \times 7 + (n - 1)2] \\ \\ \frac{n}{2}[14 + 2n - 2] \\ \\ \frac{n}{2}[2n + 12] \\ \\ \frac{n}{2} \times 2[n + 6] \\ \\ n[n + 6] \\ \\ \bold{n^2 + 6n}$

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