Question

t4=10, t10=16, t21=???
plzzzz answer it​

Answer 1

/* There is a mistake in the question , It must be like this */

$In \: A.P , t_{4} = 10 , t_{10} = 16$

$Let \: 'a' \: and \: 'd' \: are \: first \:term \: and \\common \: difference \: of \: A.P$

$\boxed{\pink{ n^{th} \: term (t_{n}) = a + (n-1)d }}$

$i ) t_{4} = 10$

$\implies a + (4-1)d = 10$

$\implies a + 3d = 10\: ---(1)$

$ii ) t_{10} = 16$

$\implies a + (10-1)d = 16$

$\implies a + 9d = 16\: ---(2)$

/* Subtract equation (1) from equation (2), we get */

$\implies 6d = 6$

$\implies d = \frac{6}{6}$

$\implies d = 1$

/* Put d = 1 in equation (1) , we get */

$a + 3 \times 1 = 10$

$\implies a + 3 = 10$

$\implies a = 10 - 3$

$\implies a = 7$

Therefore.,

$\red{ Value \: of \: t_{21} } \\= a + 20d \\= 7 + 20 \times 1 \\= 7 + 20 \\\green {= 27}$

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Answer 2

a=7

d=1

t$_{21}$=27