solve this math problem please
Answers
Answer:
Solution :-
Sum of first 10 terms = S₁₀ = 210
Using sum of n terms of an AP formula
Sₙ = n/2 (2a + (n - 1)d )
=> S₁₀ = 10/2 (2a + (10 - 1)d )
=> 210 * 2/10 = 2a + 9d
=> 42 = 2a + 9d --- eq(1)
Sum of last 15 terms S₁₅ = 2565
Sum of n terms of an AP = Sₙ = n/2(a + l)
Sum of last 15 terms of an AP = S₁₅ = 15/2(a₃₆ + a₅₀)
[ Because here, First term a = a₃₆ and last term l = a₅₀ ]
=> 2565 = 15/2 (a + 35d + a + 49d)
=> 2565 * 2/15 = 2a + 84d
=> 342 = 2a + 84d --- eq(1)
Subtracting equation (1) from equation (2)
=> 342 - 42 = 2a + 84d - 2a - 9d
=> 300 = 75d
=> 300/75 = d
=> 4 = d
Substituting d = 4 in eq(1)
=> 2a + 9d = 42
=> 2a + 9(4) = 42
=> 2a + 36 = 42
=> 2a = 42 - 36
=> 2a = 6
=> a = 6/2 = 3
General form of AP : a, a + d, a + 2d, a + 3d,....
Therefore the AP is 3, 7, 11, 15,....