Question

find ap whose nth term is given by an=9-5n.Find sum of first n terms of ap
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Answer 1

${\huge{\underline{\sf {\pink{Answer ;}}}}}$

AP = 4, -1, -6, ......... (9 - 5n)

${S}_{n} = \frac{n}{2} (13 - 5n)$

${\huge{\underline{\sf {\orange{Explanation :}}}}}$

Let,

First term be a

Last term be ${a}_{n}$

Common difference be d

Sum of n terms be ${S}_{n}$

$\: \: \: \:$

${\huge{\underline{\sf {\purple{Formula \: used :}}}}}$

${a}_{n} = a + (n - 1)d$

${S}_{n} = \frac{n}{2} \: 2a + (n - 1)d$

$\: \: \: \:$

${\huge{\underline{\sf {\blue{Solution :}}}}}$

$\: \: \: \:$

A P :

${a}_{n} = 9 -5 n \: \: \: (given)$

$\: \: \: \:$

First term, n = 1

${a}_{1} = 9 - 5(1) \: \: \\ = 9 - 5 \\ = 4$

$\: \: \: \:$

Second term, n = 2

${a}_{2} = 9 - 5(2) \\ = 9 - 10 \\ = - 1$

$\: \: \: \:$

Third term, n = 3

${a}_{3} = 9 - 5(3) \\ = 9 - 15 \\ = - 6$

$\: \:$

Common difference

= ${a}_{2} - {a}_{1}$

= (-1) - 4

= - 5

$\: \: \:$

AP = 4, -1, -6, ......... (9 - 5n)

$\: \: \:$

Sum :

${S}_{n} = \frac{n}{2} \: 2a + (n - 1)$

$\:\:\:\:\:\:= \frac{n}{2} 2(4) + (n - 1)( - 5)$

$\:\:\:\:\:\:= \frac{n}{2} 8 + (n - 1)( - 5)$

$\:\:\:\:\:\:= \frac{n}{2} \: (8 - 5n + 5 )$

$\:\:\:\:\:\:= \frac{n}{2} \: (13 - 5n)$