Question

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Answer 1

$\small\bold{\underline{\sf{\blue{\:\:Given\;\::-}}}}$

Sum of the 8th Terms is 136 and that of first 15 terms is 465.

To Find:-

Sum of first 25 Terms = ?

Explanation:-

We know that,

$\bigstar\;\:\small{\underline{\boxed{\sf{\red{\dfrac{n}{2} 2a \:+\:(n\:-\:1)d}}}}}$

Here,

a = First Term

D = Common Difference

$\bold{\underline{\sf{According\:to\:Ques\: Now}}}$

$\longrightarrow\sf\: \dfrac{8}{2} (2a \:+\:(8\:-\:1)d \:=\: 136$

$\longrightarrow\sf\: 2a + 7d = 136$ -------Eq(1)

$\rule{150}2$

$\longrightarrow\sf\: \dfrac{15}{2} (2a \:+\:(15\:-\:1)d = 465$

$\longrightarrow\sf\: 2a \:+\:14d \:=\:\dfrac{465\times\: 2}{15}$

$\longrightarrow\sf\: 2a + 14d = 62$ -------Eq(2)

$\rule{150}2$

$\dag\:\small\bold{\underline{\sf{\blue{Now,\: Subtracting\: Equation\:(1)\:from\:(2)}}}}$

$\longrightarrow\sf\: 2a \: + \: 17d \: =\: 136$

$\longrightarrow\sf\:2a \:+\: 14d \:=\: 62$

$\longrightarrow\sf\: - 7d = - 28$

$\longrightarrow\sf\: d = \dfrac{-28}{-7}$

$\longrightarrow\large\boxed{\sf{\red{d\:=\: 4}}}$

★ Substituting the Value of d in Equation (1)

$\longrightarrow\sf\: 2a + 7d = 34$

$\longrightarrow\sf\:2a + 7(4) = 34$

$\longrightarrow\sf\:2a + 28 = 34$

$\longrightarrow\sf\:2a = 34 - 28$

$\longrightarrow\sf\:2a = 6$

$\longrightarrow\sf\:a = \cancel\dfrac{6}{2}$

$\longrightarrow\large{\sf{\red{a \:=\: 3}}}$

$\rule{150}2$

Now, Finding the Sum of 25 Terms

$\longrightarrow\sf\: Sn = \dfrac{n}{2}(2a \:+\;(n \:-\:1)d$

$\longrightarrow\sf\: = \dfrac{25}{2}(6 + 24 \times \: 4)$

$\longrightarrow\sf\: \dfrac{25}{2} \times\: 102$

$\longrightarrow\large{\underline{\boxed{\sf{\pink{1275}}}}}$

$\small\bold{\underline{\sf{\blue{Hence,\;Sum\:of\: First\:25\;terms\: is\:1275}}}}$

Answer 2

Answer:

Sum of the 8th Terms is 136 and that of first 15 terms is 465.

To Find:-

Sum of first 25 Terms = ?

Explanation:-

We know that,

\bigstar\;\:\small{\underline{\boxed{\sf{\red{\dfrac{n}{2} 2a \:+\:(n\:-\:1)d}}}}}★

2

n

2a+(n−1)d

Here,

a = First Term

D = Common Difference

⟶

2

8

(2a+(8−1)d=136

\longrightarrow\sf\: 2a + 7d = 136⟶2a+7d=136 -------Eq(1)

\rule{150}2

\longrightarrow\sf\: \dfrac{15}{2} (2a \:+\:(15\:-\:1)d = 465⟶

2

15

(2a+(15−1)d=465

\longrightarrow\sf\: 2a \:+\:14d \:=\:\dfrac{465\times\: 2}{15}⟶2a+14d=

15

465×2

\longrightarrow\sf\: 2a + 14d = 62⟶2a+14d=62 -------Eq(2)

\rule{150}2

\dag\:\small\bold{\underline{\sf{\blue{Now,\: Subtracting\: Equation\:(1)\:from\:(2)}}}}†

Now,SubtractingEquation(1)from(2)

\longrightarrow\sf\: 2a \: + \: 17d \: =\: 136⟶2a+17d=136

\longrightarrow\sf\:2a \:+\: 14d \:=\: 62⟶2a+14d=62

\longrightarrow\sf\: - 7d = - 28⟶−7d=−28

\longrightarrow\sf\: d = \dfrac{-28}{-7}⟶d=

−7

−28

\longrightarrow\large\boxed{\sf{\red{d\:=\: 4}}}⟶

d=4

★ Substituting the Value of d in Equation (1)

\longrightarrow\sf\: 2a + 7d = 34⟶2a+7d=34

\longrightarrow\sf\:2a + 7(4) = 34⟶2a+7(4)=34

\longrightarrow\sf\:2a + 28 = 34⟶2a+28=34

\longrightarrow\sf\:2a = 34 - 28⟶2a=34−28

\longrightarrow\sf\:2a = 6⟶2a=6

\longrightarrow\sf\:a = \cancel\dfrac{6}{2}⟶a=

2

6

\longrightarrow\large{\sf{\red{a \:=\: 3}}}⟶a=3

\rule{150}2

Now, Finding the Sum of 25 Terms

\longrightarrow\sf\: Sn = \dfrac{n}{2}(2a \:+\;(n \:-\:1)d⟶Sn=

2

n

(2a+(n−1)d

\longrightarrow\sf\: = \dfrac{25}{2}(6 + 24 \times \: 4)⟶=

2

25

(6+24×4)

\longrightarrow\sf\: \dfrac{25}{2} \times\: 102⟶

2

25

×102

\longrightarrow\large{\underline{\boxed{\sf{\pink{1275}}}}}⟶

1275

\small\bold{\underline{\sf{\blue{Hence,\;Sum\:of\: First\:25\;terms\: is\:1275}}}}

Hence,SumofFirst25termsis1275

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