If $n^{t h}$ term of an $A P$ is $7 - 4 n$ . Find the sum of first $5$ terms.

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Solution

Given: $t_{n} = 7 - 4 n$
$\Rightarrow t_{1} = 7 - 4 \times 1 = 7 - 4 = 3$
$\Rightarrow t_{2} = 7 - 4 \times 2 = 7 - 8 = - 1$
Common difference $= d = t_{2} - t_{1} = - 1 - 3 = - 4$
Sum of first $5$ terms $= S_{5} = \frac{5}{2} [2 \times 3 + (5 - 1) (- 4)]$ using $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{5} = \frac{5}{2} [6 - 16] = \frac{5}{2} \times - 10 = - 25$

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If nth term of an AP is 7−4n. Find the sum of first 5 terms.

Given:tn=7−4n⇒t1=7−4×1=7−4=3⇒t2=7−4×2=7−8=−1Common difference=d=t2−t1=−1−3=−4Sum of first 5 terms=S5=52[2×3+(5−1)(−4)] using Sn=n2[2a+(n−1)d]S5=52[6−16]=52×−10=−25

If $n^{t h}$ term of an $A P$ is $7 - 4 n$ . Find the sum of first $5$ terms.