Question

If a.p consist of 50 term of which 3rd term is 12 and last term is 106 find 29th term

Answer 1

Hope it will help you

Answer 2

Answer:

$\bold{\sf{\therefore a_{29} =64}}$

Step-by-step explanation:

$\huge{\underline{\sf{SOLUTION-}}}$

$\implies a_{3} = 12 \\\\ \implies a + 2d = 12 - - - - - (1) \\ \\ \implies a_{50} = 106 \\ \\\implies a + 49d = 106 - - - - - (2) \\ \\ \bold{subtracting \: (1) \: from \: (2)} \\ \\ \implies a + 49d - (a + 2d) = 106 - 12 \\ \\\implies a + 49d - a - 2d = 94 \\\\ \implies 47d = 94 \\ \to d = \frac{94}{47} \\\\ { \bold{\implies d = 2}} \\ \\ \bold{putting \: value \: of \: d \: in \: (1) }\\\\ \implies a + 2d = 12 \\ \\ \implies a + 2 \times 2 = 12 \\ \\ \implies a + 4 = 12 \\ \implies a = 12 - 4 \\ \\ { \bold{\implies a = 8}}$

• We find the first term and common difference:

• So we can find a29 = ?

$\implies a_{29} = a + 28d \\\\ \implies a_{29} =8 + 28 \times 2 \\ \\ \implies a_{29} =8 + 56 \\\\ \bold{\therefore a_{29} =64}$