Question

The sum of 4th and 9th term of an ap is 46 and their product is 465. find the sum of its first 10th term.

Answer 1

Here, 
a4 + a9 = 46 
⇒ a + 3d + a + 8d = 46 
⇒ 2a + 11d = 46 
⇒ 2a = 46 – 11d 
⇒ a = 46 – 11d/2 -------------------- (1) 
Also, 
(a4)(a9) = 465 
⇒ (a + 3d)(a + 8d) = 465 
⇒ a2 + 8da + 3da + 24d2 = 465 
⇒ a2 + 11da + 24d2 = 465 
From eqⁿ (1), we get 
⇒ (46 – 11d/2)2 + 11d(46 – 11d/2) + 24d2 = 465 
⇒ 2116 – 506d – 506d + 121d2/4 + 506d – 121d2/2 + 24d2= 465 
⇒2116 – 1012d + 121d2 + 1012d – 24d2 + 96d2 = 1860 
⇒ 2116 – 121d2 + 96d2 = 1860 
⇒ 2116 – 25d2 = 1860 
⇒ 25d2 = 256 
⇒ d2 = 10.24 
⇒ d = 3.2 
From (1), 
⇒ a = 46 – 11(3.2)/2 
⇒ a = 10.8/2 
⇒ a = 5.4 
Now, Sum of first 10 terms 
⇒ S10 = 5[10.8 + (28.8)] 
⇒ 5(39.6) 
⇒198.