Question

3rd term of an AP is 15 and 10th term is 50, then sum upto 20th term is 1050. it is true or false​

Answer 1

$\huge{\blue{\bold{\underline{\underline{Answer :}}}}}$

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$\large{\green{\underline \bold{\tt{Given :-}}}}$

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• 3rd term of an AP is 15

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• 10th term is 50

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$\large{\red{\underline \bold{\tt{To \: Find :-}}}}$

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• Sum upto 20th term

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$\large{\orange{\underline{\tt{Solution :-}}}}$

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We know that

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➠ $\sf a_n = a + (n - 1)d$ ----- (1)

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Where ,

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• $\sf a_n$ = nth term

• a = First term

• d = Common difference

• n = Number of terms

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$\underline{\bold{\texttt{For 3rd term :}}}$

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• $\sf a_n$ = 15

• a = a

• d = d

• n = 3

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⟮ Putting these values in (1) ⟯

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf 15 = a + (3 - 1)d$

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➜ $\sf 15 = a + 2d$ ------ (2)

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$\underline{\bold{\texttt{For 10th term :}}}$

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• $\sf a_n$ = 50

• a = a

• d = d

• n = 10

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⟮ Putting these values in (1) ⟯

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➜ $\sf a_n = a + (n - 1)d$

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➜ $\sf 50 = a + (10 - 1)d$

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➜ $\sf 50 = a + 9d$ ----- (3)

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⟮ Subtracting equation (2) from (3) ⟯

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➜ 50 - 15 = a + 9d - (a + 2d)

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➜ 35 = a + 9d - a - 2d

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➜ 35 = 7d

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➜ $\sf d = \dfrac { 35 } { 7 }$

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➨ d = 5 ------ (4)

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• Hence common difference is 5

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⟮ Putting d = 5 from (4) to (2) ⟯

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➜ 15 = a + 2d

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➜ 15 = a + 2(5)

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➜ 15 = a + 10

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➜ a = 15 - 10

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➨ a = 5

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• Hence first term is 5

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$\underline{\bold{\texttt{Sum of terms :}}}$

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➠ $\sf S_n = \dfrac { n } { 2 } (2a + (n - 1)d)$ ---- (5)

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Where,

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• $\sf S_n$ = Sum of n terms

• n = Number of terms

• a = First term

• d = Common difference

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$\underline{\bold{\texttt{Sum of 20 terms :}}}$

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• $\sf S_n$ = $\sf S_{ 20 }$

• n = 20

• a = 5

• d = 5

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⟮ Putting these values in (5) ⟯

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

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➜ $\sf S_{ 20 } = \dfrac { 20} { 2 } (2(5)+ (20 - 1)5)$

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➜ $\sf S_{ 20 } = 10(10 + 19(5))$

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➜ $\sf S_{ 20 } = 10(10 + 95)$

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➜ $\sf S_{ 20 } = 10(105)$

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➨ $\sf S_{ 20 } = 1050$

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Thus the sum of 20 terms is 1050

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• Hence the given condition was true

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