Question

Pls answer the 25th question

Answer 1

hope it helps you out

Answer 2

From the identities we know that $a_{x}= a+(x-1)d$, so $a_{15} = a + ( 15 -1 )d = a + 14d \:\: and \:\: a_{6}= a+ (6 - 1 )d = a + 5d$

In the question,

$a_{15} - a_{6} = - \dfrac{27}{2}$

From the above explanation, we can write a₁₅ as a + ( 15 - 1 )d = a + 14d and a₆ as a + ( 6 - 1 )d = a + 5d

a₁₅ - a₆ = - $\dfrac{27}{2}$

a + 14d - ( a + 5d ) = - $\dfrac{27}{2}$

a + 14d - a - 5d = - $\dfrac{27}{2}$

a - a + 14d - 5d = - $\dfrac{27}{2}$

9d = - $\dfrac{27}{2}$

d = - $\dfrac{27}{2\times9}$

d = - $\dfrac{3}{2}$

Therefore the common difference of the arithmetic progression is - $\dfrac{3}{2}$