Let a1=8, a2, a3, ldots, an be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170 , then the product of its middle two terms is

Question

Let a1=8, a2, a3, ldots, an be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170 , then the product of its middle two terms is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

a1+a2+a3+a4=50 ⇒ 32+6 d=50 ⇒ d=3 text and, an-3+an-2+an-1+an=170 ⇒ 32+(4 n-10) ⋅ 3=170 ⇒ n =14 a7=26, a8=29 ⇒ a7 ⋅ a8=754

Q. Let $a_{1} = 8, a_{2}, a_{3}, \dots, a_{n}$ be an A.P. If the sum of its first four terms is $50$ and the sum of its last four terms is $170$ , then the product of its middle two terms is______

$a_{1} + a_{2} + a_{3} + a_{4} = 50$
$\Rightarrow 32 + 6 d = 50$
$\Rightarrow d = 3$
$and, a_{n - 3} + a_{n - 2} + a_{n - 1} + a_{n} = 170$
$\Rightarrow 32 + (4 n - 10) \cdot 3 = 170$
$\Rightarrow n = 14$
$a_{7} = 26, a_{8} = 29$
$\Rightarrow a_{7} \cdot a_{8} = 754$

Q. Let a1​=8,a2​,a3​,…,an​ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170 , then the product of its middle two terms is______

a1​+a2​+a3​+a4​=50 ⇒32+6d=50 ⇒d=3 and, an−3​+an−2​+an−1​+an​=170 ⇒32+(4n−10)⋅3=170 ⇒n=14 a7​=26,a8​=29 ⇒a7​⋅a8​=754

Q. Let $a_{1} = 8, a_{2}, a_{3}, \dots, a_{n}$ be an A.P. If the sum of its first four terms is $50$ and the sum of its last four terms is $170$ , then the product of its middle two terms is______

$a_{1} + a_{2} + a_{3} + a_{4} = 50$
$\Rightarrow 32 + 6 d = 50$
$\Rightarrow d = 3$
$and, a_{n - 3} + a_{n - 2} + a_{n - 1} + a_{n} = 170$
$\Rightarrow 32 + (4 n - 10) \cdot 3 = 170$
$\Rightarrow n = 14$
$a_{7} = 26, a_{8} = 29$
$\Rightarrow a_{7} \cdot a_{8} = 754$