the nth term of an AP is given TN=8n+3. find the sum of first n terms of the AP.
Answers
Answered by
3
Given:
Sn=8n+3.
a=S1
S2=a+a+d
=2a+d
Subtracting them we get: S2-S1=2a+d-a
=a+d =2nd term
S2-S1=a+d......................(1)
Given
Sn=8n+3
Putting n=1 and n=2 gives:-
S1=8+3
=11
a=S1=11
S2=8*2+3=16+3=19
S2-S1=19-11
=8
From(1)S2-S1=a+d
So :
a+d=8.
but a=11
so 11+d=-3
d=-3
Sn=n/2*[2a+(n-1)d]
Sn=n/2*[22-3(n-1)]
=n/2*[22-3n+3]
=n/2[25-3n]
=25n/2-3n²/2
Hope it helps you.
karinakaria:
d=8
Answered by
5
at n= 1
t1 = 8(1) +3 = 11
so, a=11
t2 = 19 = a2
so, d = 8
so, Sn. = n/2(2a+(n-1)d)
=. n/2(14+8n)
so, Sn = 7n+4n² ans.
t1 = 8(1) +3 = 11
so, a=11
t2 = 19 = a2
so, d = 8
so, Sn. = n/2(2a+(n-1)d)
=. n/2(14+8n)
so, Sn = 7n+4n² ans.
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