Question

Find the sum of 51 terms of the AP whose second term is 2 and the 4th term is 8

Answer 1

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Answer 2

Answer:

$\red { Sum \: of \: 51 \:terms (S_{51})}\green {= 3774}$

Step-by-step explanation:

$Let \: a \:and \: d \: are \: first \:term \:and \\common \: difference \: of \: an \: A.P$

$second \: term (a_{2}) = 2 \:(given )$

$\boxed { \orange { n^{th} \:term (a_{n}) = a+(n-1)d}}$

$\implies a+d = 2 \:---(1)$

$Fourth\: term (a_{4}) = 8\:(given )$

$\implies a+3d = 8\:---(2)$

/* Subtract equation (1) from equation (2) ,we get

$2d = 6$

$\implies d = \frac{6}{2} = 3$

/* Put d = 3 in equation (1) , we get

$a + 3 = 2$

$\implies a = 2- 3 = -1$

$Now, Sum \: of \: 51 \:terms (S_{51})\\= \frac{51}{2} [ 2\times (-1) + (51-1) 3]$

$\boxed {\pink { Sum \:of \:n \:terms (S_{n})=\frac{n}{2}[2a+(n-1)d]}}$

$= \frac{51}{2} [ -2+ 50\times 3 ]\\= \frac{51}{2} (-2+150)\\= \frac{51}{2} \times 148\\= 51 \times 74 \\= 3774$

Therefore.,

$\red { Sum \: of \: 51 \:terms (S_{51})}\green {= 3774}$

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