Question

${\small{\red{\mathcal{\blue{\underline{\underline{TodaY'S QuEsTiOn.}}}}}}}$

${\red{\mathbb{QUE:-}}}$

• In an AP sum of n terms = 77.
• Sum of (n-5) terms of the same AP = 7.
• The common difference of the AP iS 3.

Then Find out the $\tt n^{th}$ term. (i.e. $\tt a_n$

${\blue{\mathcal{Answer:- 20}}}$

Need explanation.

​

Answer 1

Step-by-step explanation:

d=3

(n/2)(2a+3(n-1))=77 ------(1)

Sum of (n-5) terms of the same AP = 7

therefore the sum of last 5 terms is 77-7=70

the last 5 terms are also in AP with starting number a' and common difference 3

therefore, (5/2)(2a' + 4*3)=70

solving this, a' = 8

(n-5/2)(2a+3(n-6)) = 7

(n-5/2)(2a+3(n-1))= 7 +(n-5)*15/2 ------(2)

dividing eq 1 and 2,

n/(n-5) = 77/{7+(n-5)*15/2}

here you can cross multiply and solve the quadratic in n to find the value of n(= 7)

now use this value of n in eq 1 and find a = 2

now nth term is 2 + 3*(7-1) = 20

Answer 2

$\huge\sf\pink{Answer}$

☞ Your Answer is 20

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$\huge\sf\blue{Given}$

✭ $\sf S_n = 77$

✭ $\sf S_{n-5} = 7$

✭ Common Difference (d) = 3

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

◈ The nth term?

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$\huge\sf\purple{Steps}$

We know that,

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

☯ $\underline{\sf As \ Per \ the \ Question}$

➝ $\sf S_n = \dfrac{n}{2}\bigg\lgroup 2a+(n-1)d\bigg\rgroup = 77$

➝ $\sf \dfrac{n}{2}\bigg\lgroup 2a+(n-1)3\bigg\rgroup = 77$

➝ $\sf \bigg\lgroup 2a+3n-3\bigg\rgroup = \dfrac{77\times 2}{n}$

➝ $\sf \bigg\lgroup 2a+3n-3\bigg\rgroup = \dfrac{154}{n}$

➝ $\sf 2a = \dfrac{154}{n} - 3n+3$

➝ $\sf 2a = \dfrac{154-3n^2+3n}{n}$

➝ $\sf a = \dfrac{154-3n^2+3n}{2n} \:\:\: -eq(1)$

Also given that,

➢ $\sf S_{n-5} = \dfrac{n-5}{2} \bigg\lgroup 2a+(n-5-1)3\bigg\rgroup = 7$

➢ $\sf \dfrac{n-5}{2} \bigg\lgroup 2a+(n-5-1)3\bigg\rgroup = 7$

➢ $\sf \dfrac{n-5}{2}\bigg\lgroup 2a+3n-18\bigg\rgroup = 7$

➢ $\sf 2a+3n-18 = \dfrac{7\times 2}{n-5}$

➢ $\sf 2a = \dfrac{14}{n-5}-3n+18$

➢ $\sf 2a = \dfrac{14-3n^2+15n+18n-90}{n-5}$

➢ $\sf a = \dfrac{-76-3n^2+33n}{2(n-5)}\:\:\: -eq(2)$

Comparing eq(1) & eq(2)

➠ $\sf \dfrac{154-3n^2+3n}{\cancel{2}n} = \dfrac{-76-3n^2+33n}{\cancel{2}(n-5)}$

➠ $\sf (154-3n^2+3n)(n-5) = (-76+3n^2+33n)(n)$

➠ $\sf 154n-\cancel{3n^3}+3n^2-770+15n^2-15n = -76n+\cancel{3n^3}+33n^2$

➠ $\sf 139n+18n^2-770+76n-33n^2 = 0$

➠ $\sf 215n-15n^2-770 = 0$

➠ $\sf -3n^2-43n-154 = 0$

Splitting the middle term

➳ $\sf -3n^2-21n-22n-154 = 0$

➳ $\sf -3n(n+7)-22(n+7)$

➳ $\sf (-3n-22)(n+7)$

➳ $\sf n = \dfrac{22}{3} \ or \ -7$

Ignoring Negativity and fraction

➳ $\sf \green{ n = 7}$

Substituting the value of a in eq(1)

➳ $\sf a = \dfrac{154-3n^2+3n}{2n}$

➳ $\sf a = \dfrac{154-3(7)^2+3(7)}{2(7)}$

➳ $\sf a = \dfrac{24}{14}$

➳ $\sf \red{a = 2}$

We also know that,

$\underline{\boxed{\sf a_n = a+(n-1)d}}$

»» $\sf a_7 = 2+(7-1)3$

»» $\sf a_7 = 2+(6)(3)$

»» $\sf a_7 = 2+18$

»» $\sf \orange{a_7 = 20}$

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