Given sum of first n terms of an AP is $$2n$$ $$+$$ $$3n2$$. Another AP is formed with the same first term and double of the common difference, the sum of n terms of the new AP is:

Question

Infinity Learn · Accepted Answer

Given sum of n terms of AP is $$Sn=2n+3n2$$ For the first term substitute $$n=1$$ Hence, $$a1$$ $$=$$ $$5$$ The second term is $$a2=S2-S1=16-5=11$$ The common difference is $$a2-a1=11-5=6$$ Hence, the first term and common difference of the original sequence are $$a1$$ $$=5$$, $$d=6For$$ the new sequnce, first term is same $$a=5$$ only  Common difference is two times of $$6d=12Sum$$ of first n terms of new AP is $$Sn=n22a+n-1dSn=n210+(n-1)12$$ $$=6nn-n$$ $$=6n2-n$$

Given sum of first n terms of an AP is 2n + 3n². Another AP is formed with the same first term and double of the common difference, the sum of n terms of the new AP is:

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Given sum of first n terms of an AP is 2n + 3n2. Another AP is formed with the same first term and double of the common difference, the sum of n terms of the new AP is:

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Given sum of first n terms of an AP is 2n + 3n². Another AP is formed with the same first term and double of the common difference, the sum of n terms of the new AP is: