Question

Sum of the first n terms of an arithmetic progression is n^2+4n. Find the 15th term of the progression
please solve it....​

Answer 1

Answer:

$\huge{\pink{\green{\mathsf{a15=33}}}}$

Step-by-step explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

$\: \: \: \: \: \: { \orange{given}} \\ { \green{Sn = {n}^{2} + 4n }} \\ { \green{let \: n = 1,2,3.....}} \\ \\ { \blue{to \: find}} \\ { \red{a15 =? }}$

☆ According to given question:

$\to Sn = {n}^{2} + 4n \\ \to S1 = {1}^{2} + 4 \times 1 \\ \to S1 = 1 + 4 \\ \to S1 = 5 = a1\\ \\ \to S2 = {2}^{2} + 4 \times 2 \\ \to S2 = 4 + 8 \\ \to S2 = 12 \\ \\ \to S3 = {3}^{2} + 4 \times 3 \\ \to \: S3 = 9 + 12 \\ \to S3 = 21 \\ \\ \to a2 = S2 - S1 \\ \to a2 = 12 - 5 \\ \to a2 = 7 \\ \\ \to a3 = S3 - S2 \\ \to a3 = 21 - 12 \\ \to a3 = 9 \\ \\ \to d = a2 - a1 \\ \to d = 7 - 5 \\ \to d = 2$

☆ first term=5

☆ common difference=2

$\to a15 = a + 14d \\ \to a15 = 5 + 14 \times 2 \\ \to a15 = 5 + 28 \\ { \green{\to a15 = 33}}$

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