if the sum of the first k terms of an ap is 3k-k and its common differenceis 6,then find the first term?
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Sum of K terms is Sk.
d is common difference.
a1 is the first term of AP.
ak is the k th term of AP.
sum of first term is S1.
Sk=1/2 (3k*2+7k)
k/2 [2a+(k-1)d]=1/2 (3k*2+7k)
k [2a+(k-1)d]=3k*2+7k
2a+(k-1)d=3k+7
a+1/2 (k-1)d=3k+7................ (1)
let a+1/2(k-1)d=k th term
a+1/2 (k-1)d=a+19d
by comparison,
(k-1)÷2=19
k=39..................... (2)
If we place 39 in place of 'k' in equation (1)
then,
we have
a20=(3×39+7)÷1/2
a20=62************
S1=a1
1/2 (3×1*2+7×1)=a ( first term of AP)
a=5........................(3)
a20-a1=62-5
19d=57
d=3.......................... (4)
So,
ak=a+(k-1)×3
ak=5+(k-1)×3
ak=5+3k-3
ak=3k+2************
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d is common difference.
a1 is the first term of AP.
ak is the k th term of AP.
sum of first term is S1.
Sk=1/2 (3k*2+7k)
k/2 [2a+(k-1)d]=1/2 (3k*2+7k)
k [2a+(k-1)d]=3k*2+7k
2a+(k-1)d=3k+7
a+1/2 (k-1)d=3k+7................ (1)
let a+1/2(k-1)d=k th term
a+1/2 (k-1)d=a+19d
by comparison,
(k-1)÷2=19
k=39..................... (2)
If we place 39 in place of 'k' in equation (1)
then,
we have
a20=(3×39+7)÷1/2
a20=62************
S1=a1
1/2 (3×1*2+7×1)=a ( first term of AP)
a=5........................(3)
a20-a1=62-5
19d=57
d=3.......................... (4)
So,
ak=a+(k-1)×3
ak=5+(k-1)×3
ak=5+3k-3
ak=3k+2************
Read more on Brainly.in - https://brainly.in/question/2188226#readmore
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