Question

Find the sums given below :

1. $7 + 10\frac 12+ 14 + ....... + 84$
2. $34 + 32 + 30 + ....... + 10$
3. $–5 + (–8) + (–11) + ....... + (–230)$

Answer 1

(i) $7 + 10\frac 12 + 14 + …………+ 84$
For this A.P.,
$a = 7$ and $l = 84$
$d = a_2-a_1 = 10\frac 12 -7 = \frac {21}{2} - 7 = \frac 72$
Let 84 be the nth term of this A.P.
$l = a + (n − 1)d$
$84 = 7 + (n-1)\frac 72$
$77 = (n-1)\frac 72$
$22 = n − 1$
$n = 23$
We know that,
$S_n = \frac n2[a+l]$

$S_{23} = \frac {23}{2}[7+84]$

$S_{23} = \frac {23 \times 91}{2}$

$S_{23} =\frac {2093}{2}$

$S_{23} = 1046\frac 12$

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(ii) $34 + 32 + 30 + ……….. + 10$
For this A.P., $a = 34$, $d = a_2 − a_1 = 32 − 34 = −2$ and $l = 10$
Let $10$ be the n^th term of this A.P. 
$l = a + (n − 1) d$
$10 = 34 + (n − 1) (−2)$
$−24 = (n − 1) (−2)$
$12 = n − 1$
$n = 13$
$S_n = \frac n2[a+l]$

$S_n = \frac {13}{2}[34+10]$

$Sn = \frac {13 \times 44}{2}$
$S_n = 13 \times 22$
$S_n = 286$

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(iii) $(−5) + (−8) + (−11) + ………… + (−230)$
For this A.P.,
$a = −5$, $l = −230$ and $d = a_2 − a_1 = (−8) − (−5) = − 8 + 5 = −3$
Let $−230$ be the n^th term of this A.P.
$l = a + (n − 1)d$
$−230 = − 5 + (n − 1) (−3)$
$−225 = (n − 1) (−3)$
$n − 1 = 75$
$n = 76$
And, $S_n =\frac n2[a+l]$

$S_n = \frac {76}{2}[(-5)+(-230)]$
$S_n = 38(-235)$
$S_n = - 8930$