Let there be an A.P with first term $a$ , common difference $d$ . If $a_{n}$ denotes its $n$ th term and $S_{n}$ the sum of first $n$ terms, , if $a_{n} = 4, d = 2$ and $S_{n} = - 14$ then $| n + a |$ is

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Solution

Given,

$d = 2$

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

$4 = a + (n - 1) 2$

$4 = a + 2 n - 2$

$a + 2 n = 6$

$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$- 14 = n [a + n - 1]$

$- 14 = n [6 - 2 n + n - 1]$
$- 14 = n [5 - n]$
$14 = n [n - 5]$
$7 [7 - 5] = n [n - 5]$
$∴ n = 7$

$A n d, a + 2 (7) = 6 \Rightarrow a = - 8$

$∴ | a + n | = | - 8 + 7 | = | - 1 | = 1$

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Let there be an A.P with first term a, common difference d. If an denotes its nth term and Sn the sum of first n terms, , if an=4,d=2 and Sn=−14 then|n+a| is

Given,d=2an=a+(n−1)d4=a+(n−1)24=a+2n−2a+2n=6Sn=n2[2a+(n−1)d]−14=n[a+n−1]−14=n[6−2n+n−1]−14=n[5−n] 14=n[n−5] 7[7−5]=n[n−5]∴n=7And,a+2(7)=6⇒a=−8∴|a+n|=|−8+7|=|−1|=1

Let there be an A.P with first term $a$ , common difference $d$ . If $a_{n}$ denotes its $n$ th term and $S_{n}$ the sum of first $n$ terms, , if $a_{n} = 4, d = 2$ and $S_{n} = - 14$ then $| n + a |$ is