Question

The 11th term of an AP is -7 while the 3rd term is 9,find the first four terms

Answer 1

Answer:

$\large{\underline{\boxed{\sf 13,11,9,7...}}}$

Step-by-step explanation:

Let the terms of A.P be a, a + d, a + 2d, a + 3d...

Given that ;

a₁₁ = - 7

a₃ = 9

We know that ;

According to the question ;

$\sf a_{11} = - 7 \\ \\ \implies \sf a + 10d = - 7 \\ \\ \implies \sf a = - 7 - 10d \: \: \: ....(i)$

Also,

$\sf a_3 = 9 \\ \\ \implies \sf a + 2d = 9 \\ \\ \implies \sf a = 9 - 2d \: \: \: ....(ii)$

Now, on comparing eqⁿ (i) and (ii),

$\implies \sf 9 - 2d = - 7 - 10d \\ \\ \implies \sf - 2d + 10d = - 7 - 9 \\ \\ \implies \sf 8d = - 16 \\ \\ \implies \sf d = - \frac{16}{8} \\ \\ \implies \sf d = - 2$

Putting the value of d in (ii),

$\implies \sf a = 9 - 2d \\ \\ \implies \sf a = 9 - 2( - 2) \\ \\ \implies \sf a = 9 + 4 \\ \\ \implies \sf a = 13$

Thus, the terms of A.P are;

• a₁ = 13
• a₂ = 13 + (-2) = 11
• a₃ = 13 + 2(-2) = 9
• a₄ = 13 + 3(-2) = 7

Hence, the first four terms of A.P are 13, 11, 9, 7....