Question

The sum of 15 terms of A. P 3,6,9,..

Answer 1

Answer:

We will now evaluate the sum of the first 15 terms of AP. We know that the sum of first n terms of AP is given by Sn=n2[2a+(n−1)d]. Substituting n=15,a=10,d=4 in the above equation, we have S15=152[2(10)+(15−1)4]. Simplifying the above expression, we have S15=570

Answer 2

★ Question :-

• Find the sum of first 15 terms, if the A.P = 3 , 6 , 9 , ....

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★ Answer :-

• The sum of first 15 term = 360

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★ Explanation :-

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✪ Given :-

• A = 3
• D = 3
• N = 15

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✪ To Find :-

• The sum of first 15 terms.

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✪ Solution :-

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⠀⠀ $\maltese\:\tt\large\pink{As\:We\:Know}\:\maltese\\ \\\sf{S\:=\frac{n}{2}\:[\:2\:a+(n-1)\:d\:]}\\ \\\maltese\:\tt\large\pink{Now,\:Lets\: Substitute}\:\maltese\\ \\ \tt{S = \frac{15}{2}\:[\:2×3+(15-1)×3\:]}\\ \\ \tt{S=\frac{15}{2}\:[\:2×3+(\:14\:)×3\:]} \\ \\ \tt{S=\frac{15}{2}[\:6+(\:42\:)\:]} \\ \\ \tt{S=\frac{15}{\cancel{2}}[\:\cancel{48}\:]} \\ \\ \tt{S = 15×24} \\ \\ \boxed{\bold\red{S=360}}$

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Q.] The sum of 15 terms of A.P 3, 6, 9, …..

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