Question

In arithmetic progression, the fifth term exceeds the third terms by 40. The sum of the forst ten terms is 2100, if the last terms is 409. Find the sum of all terms in sequence

Answer 1

sum of all terms in sequence = 3900

Step-by-step explanation:

Correction Last terms is 400

5th term = a + 4d

3rd Term = a + 2d

the fifth term exceeds the third terms by 40

5th term = third term + 40

=> a + 4d = a +2d + 40

=> 2d = 40

=> d = 20

Sum of 1st 10 term = 2100

(10/2)(a + a + 9d)  = 2100

=> 2a + 9d = 420

=> 2a + 9*20 = 420

=> 2a = 240

=> a = 120

Last term = a + (n-1)d  = 400

=> 120 + (n - 1)20 = 400

=> (n - 1) = 280/20

=> n - 1 = 14

=> n = 15

Sum of all terms = (15/2)(120 +  400)  = 3900

sum of all terms in sequence = 3900

Sequence : 120 , 140 , 160 ......................................, 380 , 400

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