Question

Sn=n(2n-1) find series
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Answer 1

Answer:

The series is 1,6,15,28,45,66.....

Step-by-step explanation:

This answer is correct please thank me

Answer 2

Answer:

Question:-

$\bf \:If \: S_n = n(2n - 1), \: find \: an \: ap \: series.$

$\bf\underbrace\orange{Answer:}$

Given :-

$\bf \:S_n = n(2n - 1)$

To Find :-

• An AP series.

Formula used :-

$\bf \:S_n = \dfrac{n}{2} (2a + (n - 1)d)$

where,

• a = first term of AP
• d = Common Difference of an AP
• n = number of terms of AP

$\bf\underbrace\orange{Solution:}$

Let assume that first term of an AP be 'a' and common difference be 'd' having number of terms 'n'.

According to statement

$\bf \:S_n = n(2n - 1)$

$\bf\implies \: \dfrac{n}{2} (2a + (n - 1)d) = n(2n - 1)$

On cancelation of n on both sides, we get

$\bf\implies \:\dfrac{1}{2} (2a + (n - 1)d) = 2n - 1$

$\bf\implies \:2a + nd - d = 4n - 2$

$\bf\implies \:(2a - d)+ nd = 4n - 2$

On comparing, we get

$\bf\implies \:d = 4\: and \: 2a - d = - 2$

$\bf\implies \:2a - 4 = - 2$

$\bf\implies \:2a = 2$

$\bf\implies \:a = 1$

$\bf \:So, \: required \: AP \: series \: is \: 1,5,9,13,.....$

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