Question

the sum of the first 9 terms is 81 and the sum of first 20 terms is 200. what is the first term and the common difference?

Answer 1

$\sf{\underline{\boxed{\large{\blue{\mathsf{Correct\: Question}}}}}}$

• the sum of the first 9 terms is 81 and the sum of first 20 terms is 400. what is the first term and the common difference?

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$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• Sum of first 9 terms = 81
• sum of first 20 terms = 400

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• common diff = d = ??
• first term = a= ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

$\sf{\red{\boxed{\bold{S = \dfrac{n}{2}[2a + ( n - 1 ) d ]}}}}$

☆:$\sf\implies 81 = \dfrac{9}{2}[ 2a + ( 9 - 1 ) d ]$

:$\sf\implies 81 = \dfrac{9}{2}[ 2a + 8 d ]$

:$\sf\implies 81 \times \dfrac{2}{9} = [ 2a + 8d ]$

:$\sf\implies 18 = [ 2a + 8d ]$

$\sf{\bold{\orange{\underline{\underline{2a + 8d = 18 }}}}}$------- ( i )

☆:$\sf\implies 400 = \dfrac{20}{2}[ 2a + ( 20 - 1 ) d ]$

:$\sf\implies 400 = 10 [ 2a + 19 d ]$

:$\sf\implies 400 \times \dfrac{1}{10} = [ 2a + 19d ]$

:$\sf\implies 40 = [ 2a + 19d ]$

$\sf{\bold{\orange{\underline{\underline{2a + 19d = 40 }}}}}$------- ( ii )

• Subtract eq i from ii

$\sf\longrightarrow 2a + 19d - 2a - 8d = 40 - 18$

$\sf\longrightarrow 11d = 22$

$\sf\longrightarrow d = 2$

$\sf{\blue{\boxed{\bold{common \: difference = 2 }}}}$

• putting value of d in eq i

$\sf\longrightarrow 2a + 8d = 18$

$\sf\longrightarrow 2a + 8\times 2 = 18$

$\sf\longrightarrow 2a + 16 = 18$

$\sf\longrightarrow 2a = 18 - 16$

$\sf\longrightarrow 2a = 2$

$\sf\longrightarrow a = \dfrac{2}{2}$

$\sf\longrightarrow a = 1$

$\sf{\blue{\boxed{\bold{a = 1 }}}}$

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• common difference = 2
• first term = 1

Answer 2

Answer:

thanks for the free point my friend