Question

Find the sum of n terms of an A.P. whose nth term is given by an=5-6n

Answer 1

$\huge{ \mathtt{Given-}}$

an=5-6n

$\huge{ \mathtt{\purple{SOLUTION:-}}}$

$\large{ \bold{Putting \: n=1 }}$

$\large{=>t1=-1}$

$\large{ \bold{Putting \: n=3}}$

$\large{=>t3=-3}$

$\small{\boxed{ \bold{ \blue{Sum \: of \: n \: terms = \frac{n}{2} (2a + (n - 1) - 6}}}}$

$\large{ \bold{ = > \frac{n}{2} (2 ( - 1) + (n - 1) - 6)}}$

$\large{ \bold{ = > \frac{n}{2} ( - 2 - 6n + 6)}}$

$\large{ \bold{ = > \frac{n}{2} (4 - 6n)}}$

$\large{ \bold{ = > n(2 - 3n)}}$

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$\large{ \mathtt{ \red{Thanks...♡}}}$

Answer 2

Answer:

given: tn = 5 - 6n

Put n=1, t1= -1

n=2, t2= -7

n=3,t3= -13

Sum of n terms= n/2[2a+(n-1)d]

= n/2[2(-1)+(n-1)-6]

= n/2[-2-6n+6]

=n/2(4-6n)

=> n(2-3n)