Question

In a A.P. the 11th term is 16 and 21st term is 29 then find 41st term of an A.P.

Answer 1

so a41= 31. tis is the answer

Answer 2

Answer:

$tn = 16 \: and \: t21 = 29 \\ let \: a \: be \: the \: first \: term \: and \: d \: the \: common \: difference \\ tn = a + (n - 1)d \\ t11 = a + (11 - 1)d \\ 16 = a + 20d \\ subtracting \: equation \: (1)from \: equation(2) \\ 29 = a + 20d.....(2) \\ 16 = a + 10d.....(1) \\ 13 = \: \: \: \: \: \: \: 10d \\ \: \: d = 13 \div 10 = 1.3 \\ substituting \: d = 1.3 \: in \: equation(1) \\ 16 = a + 10 \times 1.3 \\ 16 = a + 13 \\ a = 16 - 13 \\ a = 3 \\ now \: t41 = a + (41 - 1)d \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 3 + 40 \times 1.3 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 3 + 52 \\ \: \: \: \: \: t41 \: = 55 \\ the \: 41st \: term \: of \: this \: a.p \: is \: 55$

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