Math, asked by jagdishjambe236, 11 months ago

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

Answers

Answered by Anonymous
43

 \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)}}

━━━━━━━━━━━━━━━━━━━━━━━━━━

 \large{ \mathtt{ \red{Thanks...♡}}}

Answered by Anonymous
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)

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