If the sum of three consecutive terms of an increasing ap is 99 and the product of the first and third of these terms is 1025 then find the third term
Answers
GIVEN :
Sum of three consecutive terms of an increasing AP = 99
Let the terms be a - d , a , a + d
(a - d) + a + a + d = 99
a - d + a + a + d = 99
3a = 99
a = 99/3
a = 33
Product of first and third terms = 1025
( a - d ) × ( a + d ) = 1025
a² - d² = 1025 [ (a - b) (a + b) = a² - b² ]
33² - d² = 1025
1089 - d² = 1025
-d² = 1025 - 1089
-d² = -64
d² = 64
d = √64
d = 8
Third Term = a + d = 33 + 8 = 41
Therefore, the third term is 41.
The sum of three consecutive terms of an AP is 99.
• Let three consecutive terms be a - d, a, a + d.
And the sum of these consecutive terms is 99.
=> a - d + a + a + d = 99
=> 3a = 99
=> a =
=> a = 33 ________ (eq 1)
_______________________________
The product of first and third term of consecutive terms of an AP is 1025
» First term = a - d
» Third term = a + d
- A.T.Q.
=> (a - d) (a + d) = 1025
We know that (a - b) (a + b) = a² - b²
=> a² - d² = 1025
=> (33)² - d² = 1025 _____ [From (eq 1)]
=> 1089 - d² = 1025
=> - d² = 1025 - 1089
=> - d² = -64
=> d² = 64
=> d = √64
=> d = 8
_______________________________
We have to find the third term i.e. a + d
We have a = 33 and d = 8
› Take a = 33 and d = + 8
=> a + d = 33 + 8
=> a + d = 41
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The third term is 41.
______________ [ANSWER]