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 _________ .        [2023]


Ans.

(754)

Sum of first four terms =42[16+3d]

⇒50=32+6d⇒d=3

Now, sum of last four terms = 170

⇒42[2(8+(n-4)d)+3d]=170             [∵ last four terms are an-3, an-2, an-1, an]

⇒2[2(8+(n-4)·3)+9]=170

⇒16+6(n-4)+9=85

⇒6(n-4)=60⇒n-4=10⇒n=14

∴  Middle terms are 7th term and 8th term.

∴  T7=8+6d=8+18=26,   T8=8+7d=8+21=29

Now, T7×T8=26×29=754