If S1,S2,S3.............Sr are the sums of first n terms of r arithmetic progressions whose first terms are 1,2,3........ and whose common difference are 1,3,5..........respectively, then the value of S1+S2+S3+..................+Sr is _______________. a) (nr-1)(nr+1) /2 b) (nr+1)nr /2 c) (nr-1)nr /2 d) n(nr+1) /2 ___________________________________________ Please show the method for solving this problem.... THANKUU....!!
Answers
Solution :-
→ First Term = a = 1,2,3, ___________ r .
→ common Difference = d = 1,3,5,________ (2r - 1) .
→ n = n
we know that, sum of n term of an AP , with first term as a and common difference as d is :- (n/2)[2a + (n - 1)d] .
So,
→ S1 = (n/2)[2*1 + (n - 1)1] = (n/2)[ 2 + (n -1)]
→ S2 = (n/2)[2*2 + (n - 1)3] = (n/2)[ 4 + (n - 1)3]
→ S3 = (n/2)[2*3 + (n - 1)5] = (n/2)[ 6 + (n - 1)5]
Similarly,
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→ Sr = (n/2)[2*r + (n - 1)(2r - 1)] = (n/2)[ 2r + (n - 1)(2r - 1)]
Now, we have to Find :-
→ S1 + S2 + S3 + _________________ + Sr
Putting values Now, we get,
→ (n/2)[ 2 + (n - 1)] + (n/2)[ 4 + (n - 1)3] + (n/2)[ 6 + (n - 1)5] + _____________ + (n/2)[ 2r + (n - 1)(2r - 1)]
→ (n/2)[2 + 4 + 6 + __________ 2r] + (n/2)(n - 1)[ 1 + 3 + 5 + ____ (2r - 1)]
Now as we can see , both inside parts are also in AP .
Solving them ,
→ (2 + 4 + 6 + _______ 2r) = (r/2)[2*2 + (r - 1)2] = (r/2)[ 4 + 2r - 2] = (r/2)[ 2r + 2 ] = (r/2) * 2 * [r + 1] = r(r + 1)
and,
→ 1 + 3 + 5 + ____ (2r - 1) = (r/2)[2*1 + (r - 1)2] = (r/2)[ 2 + 2r - 2 ] = (r/2) * 2r = r²
Therefore,
→ (n/2)[r(r + 1)] + (n/2)(n - 1)[r²]
→ (n/2) * r. [ (r + 1) + (n - 1) * r ]
→ (nr/2) [ r + 1 + nr - r ]
→ (nr/2)[ nr + 1 ] (b) (Ans.)