The sum of all terms of …
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Gaurav Seth 5 years, 4 months ago
First term is a and common difference is d.
Sum of first and third terms is 10.
t₁ + t₅ = 10
==> a + (1 - 1) * d + a + (5 - 1) * d = 10
==> a + a + 4d = 10
==> 2a + 4d = 10
==> a + 2d = 5 ---- (1)
Ten Terms except for the first term is 99 and except for sixth term is 89.
S₁₀ - First term = 99
==> 10/2[2a + (10 - 1) * d] - a = 99
==> 5[2a + 9d] - a = 99
==> 10a + 45d - a = 99
==> 9a + 45d = 99
==> a + 5d = 11 ------- (2)
Subtract Equation (1) and (2).
a + 5d - a - 2d = 11 - 5
==> 3d = 6
==> d = 2
Place d = 2 in (2).
==> a + 5d = 11
==> a + 10 = 11
==> a = 1
Third term of the progression.
t₃ = 1 + (3 - 1) * 2
t₃ = 5
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