Question

In an AP , Given a12 = 37,d=3 find a and s12​

Answer 1

$\sf \: a = 4 \: and \: {s}^{12} = 246$

Given :-

• $\sf \: {a}^{12} = 37$
• $\sf \: d = 3$
• $\sf \: n = 12$

To Find :-

• $\sf \: Value \: of \: a \: and \: {s}^{12}$

Solution :-

${ \underline{According \: to \: the \: question, }}$

$: \implies \: \: \: \sf \: a_{n} = a + (n - 1)d \\ : \implies \: \: \: \sf \: 37 = a + (12 - 1)3 \\ : \implies \: \: \: \sf \: 37 = a + (11)3 \\ : \implies \: \: \: \sf \: 37 = a + 33 \\ : \implies \: \: \: \sf \: a = 37 - 33 \\ : \implies \: \: \: { \boxed{ \blue{ \sf \: a = 4}}}$

Now,

$: \implies \: \: \: \sf \: s_{n} = \dfrac{n}{2} (a + 1) \\ : \implies \: \: \: \sf \: s_{n} = \dfrac{12}{2}(4 + 37) \\ : \implies \: \: \: \sf \: s_{n} = 6 \times 41 \\ : \implies \: \: \: \: { \boxed{ \blue{ \sf \: s_{n} = 246}}}$