Question

For an A.P., t4=12 and its common difference d=-10, then find tn.​

Answer 1

Answer:

a+(4-1)d = 12

a+3(10) = 12

a+30 = 12

a = -18

an = (-18)+(n-1)(10)

an = -18+10n-10

an = -28+10n

Answer 2

$\sf \: Question:For \: an \: A.P., \: t_{4} =12 \: and \: \\ \sf \: its \: common \: difference \: d=(-10), \\ \sf then \: find \: t_{n}$

$\large \bf \: Given: \\ \rightarrow \sf t_{4} = 12 \\ \rightarrow \sf d = ( - 10)$

$\large \bf To \: Find : \\ \hookrightarrow \sf \: t_{n}$

$\large \bf Solution: \\ \rightarrow \sf t_{4} = a + (4 - 1)d = 12 \\ \sf a + 3( - 10) = \sf 12 \\ \sf a - 30 = 12 \\ \boxed{ \sf a = 42}$

$\sf Now \\ \sf t_{n} = a + (n - 1)d \\ \sf = 42 + (n - 1)( - 10) \\ \boxed{ \boxed { \sf t_{n} = 52 - 10n}}$

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