Find the sums given below:
(i) $7 + 10\frac{1}{2} + 14 + … + 84$
(ii) $34 + 32 + 30 + … + 10$
(iii) $-5 + (-8) + (-11) + ….. + (-230)$

To do:

We have to find the given sums.

Solution:

(i) $a=7, d=10\frac{1}{2}-\frac{7}{1}=\frac{21}{2}-\frac{7}{1}=\frac{21-14}{2}=\frac{7}{2}, a_n=84$

We know that,

$a_n=a+(n-1)d$

$S_{n}=\frac{n}{2}[a+l]$

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

$84=7+(n-1) \frac{7}{2}$

$84-7=(n-1) \frac{7}{2}$

$77(\frac{2}{7})=n-1$

$11(2)=n-1$

$22+1=n$

$n=23$

Therefore,

$S_{n}=\frac{n}{2}[a+l]$

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

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

$=\frac{2093}{2}$

$=1046\frac{1}{2}$

(ii) $a=34, d=32-34=-2, a_n=10$

We know that,

$a_n=l=a+(n-1)d$

$S_{n}=\frac{n}{2}[a+l]$

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

$10=34+(n-1)(-2)$

$10-34=(n-1)(-2)$

$\frac{-24}{-2}=n-1$

$n-1=12$

$n=12+1$

$n=13$

Therefore,

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

$=\frac{13}{2}(44)$

$=13 \times 22$

$=286$

(iii) Here,

$(-5)+(-8)+(-11)+\ldots+(-230)$ is in A.P.

$a=-5, d=-8-(-5)=-8+5=-3$ and $l=-230$

We know that,

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

$\Rightarrow-230=-5+(n-1)(-3)$

$-230=-5-3 n+3$

$3 n=-5+3+230=228$

$n=\frac{228}{3}=76$

$S_{n}=\frac{n}{2}[a+l]$

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

$=38(-5-230)$

$=38(-235)$

$=-8930$

Therefore, the sum of the given sequence is $-8930$.  

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Updated on: 10-Oct-2022

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