Question

Consider arithmetic sequence 11, 15, 19...327. Can the sum of any 25 terms of this sequence 2020? Why?​

Answer 1

$\huge\sf\pink{Answer}$

☞ No it is not possible

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$\huge\sf\blue{Given}$

✭ AP - 11,15,19....327

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$\huge\sf\gray{To \:Find}$

◈ If the sum of 25 terms can be 2020?

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$\huge\sf\purple{Steps}$

We know that,

$\underline{\boxed{\sf S_n = \dfrac{n}{2}\bigg\lgroup 2a+(n-1)d\bigg\rgroup}}$

Where,

$\qquad$✦ n = the number of the term in the AP = 25

$\qquad$✦ a = First term = 11

$\qquad$✦ d = Common difference = 4

Substituting the given values,

➳ $\sf S_{25} = \dfrac{25}{2} \bigg\lgroup 2(11)+(25-1)(4)\bigg\rgroup$

➳ $\sf S_{25} = \dfrac{25}{2} \bigg\lgroup 22+(24)(4)\bigg\rgroup$

➳ $\sf S_{25} = 25\bigg\lgroup 11 + 96\bigg\rgroup$

➳ $\sf S_{25} = 25\bigg\lgroup 107\bigg\rgroup$

➳ $\sf \red{S_{25} = 2627}$

But,

➝ $\sf\orange{2627\ne 2020}$

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