Question

In an AP given an = 4, d = 2, Sn = -14 find n and a​

Answer 1

Answer:

an=a (n-1)d

putting values:

4=a (n-1) 2

a(n-1)=2

a(n) =2-1

a(n) = 1

a =1 and n= 1

sn is used in question of finding sum

Answer 2

Given :

• $\sf \: a_{n} = 4$
• $\sf \: d = 2$
• $\sf \: S _{n} = - 14$

Find :

• Find the value of n and a .

Solution :

$\boxed{ \red{\sf \: a_{n} = a + (n - 1)d }}\\ \\ \rightarrow \sf4 = a + (n - 1)2 \\ \rightarrow \sf4 = a + 2n -2 \\ \rightarrow \sf6 = a + 2n \\ \boxed{ a = 6 - 2n} \\ \\ \: \: \boxed{ \red{\sf S_{n} = \frac{n}{2} (2a + (n - 1)d)}} \\ \\ \rightarrow \sf \: - 14 = \frac{n}{2}[ 2 (6n - 2n) + (n - 1)2] \\ \\ \rightarrow \sf - 14 = \frac{n}{2} (12 - 4n + 2n - 2) \\ \\ \rightarrow \sf - 14 = \frac{n}{2} ( 10_{n} - 2_{n}) \\ \\ \rightarrow \sf - 28 = n(10 - 2n) \\ \\ \rightarrow \sf - 28 = 10n - 2n {}^{2} \\ \\ \rightarrow \sf2n {}^{2} - 28 - 10n = 0 \\ \\ \implies \sf \: 2n {}^{2} - 10n - 28 = 0 \\ \\ \sf\colorbox{orange}{All these terms divisible by 2} \\ \\ \rightarrow \sf n {}^{2} - 5n - 14 = 0 \\ \\ \rightarrow \sf(n - 7)(n + 2) \\ \\ \boxed{ \sf \: n - 7 = 0 \implies \: n = 7} \\ \\ \boxed{ \sf \: n + 2 \implies \: n = - 2} \\ ( \sf\green{number \: of \: terms \: can \: not \: be \: negative)} \\ \sf \: so \: the \: value \: of \: n \: is \: \\ \boxed{ \green {\sf \: n =7 }} \\ \\ \boxed{ \sf \: a = 6 - 2n} \\ \rightarrow \: \sf a = 6 - 2 \times 7 \\ : \implies \sf \: a = 6 - 14 \\ : \implies \boxed{ \green{ \sf \: a = - 8}}$

Therefore the value of a = -8 and value of n = 7 .

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