Question

the sum of first n terms of an ap whose 1st term is 8 and common difference is 20, is equal to the sum of first 2n term of another AP whose 1st term is -30 and common difference is 8 ,find n.​

Answer 1

GIVEN :–

The sum of first n terms of an A.P. whose 1st term is 8 and common difference is 20, is equal to the sum of first 2n term of another AP whose 1st term is -30 and common difference is 8.

TO FIND :–

• Value of 'n' = ?

SOLUTION :–

• If 'a' is first term and 'd' is Common difference then sum –

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

• According to the first condition –

$\\ \: \implies \: \sf S_1 = \dfrac{n}{2}[2(8)+ (n - 1)(20)]$

$\\ \: \implies \: \sf S_1 = \dfrac{n}{2}[16+ 20n - 20]$

$\\ \: \implies \: \sf S_1 = \dfrac{n}{2}[20n - 4]$

$\\ \: \implies \: \sf S_1 = n(10n - 2)$

$\\ \: \implies \: \sf S_1 = 10{n}^{2} - 2n \: \: \: - - - eq.(1)$

• According to the second condition –

$\\ \: \implies \: \sf S_2 = \dfrac{2n}{2}[2( - 30)+ (2n - 1)(8)]$

$\\ \: \implies \: \sf S_2 = n[ - 60+ 16n - 8]$

$\\ \: \implies \: \sf S_2 = n[16n - 68]$

$\\ \: \implies \: \sf S_2 = 16{n}^{2} - 68n \: \: \: - - - eq.(2)$

• Now According to the question –

$\\ \: \implies \: \sf S_1 = S_2$

$\\ \: \implies \: \sf 10{n}^{2} - 2n = 16{n}^{2} - 68n$

$\\ \: \implies \: \sf 10{n}^{2} - 16{n}^{2} = 2n - 68n$

$\\ \: \implies \: \sf -6{n}^{2}= - 66n$

$\\ \: \implies \: \sf -6{n}^{2} + 66n = 0$

$\\ \: \implies \: \sf n(n - 11) = 0$

$\\ \: \implies \: \sf n = 0 \: , \: n = 11$

Answer 2

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