The sum of the first 25 terms of an A.P. is 1700, while its common dif

1.	The sum of the first 25 terms of an A.P. is 1700, while its common difference is 6. What is the 31st term of the10.A.P.?
Answer» Answer: Step-by-step explanation: Sₙ = n/2[2A + (n-1)d] 1700 = 25/2[2a + 24d] => 2a + 24d = 1700 * 2/25 => 2a + 24d = 136 => 2a + 24*6 = 136 => 2a = 136 - 144 => 2a = -8 => a = -4. nth TERM of AP is given by tₙ = a + (n-1)d 31st term is t₃₁ = -4 + (31 - 1)6 = - 4 + 180 = 176.

The sum of the first 25 terms of an A.P. is 1700, while its common difference is 6. What is the 31st term of the10.A.P.?

Answer»

Answer:

Step-by-step explanation:

Sₙ = n/2[2A + (n-1)d]

1700 = 25/2[2a + 24d]

=> 2a + 24d = 1700 * 2/25

=> 2a + 24d = 136

=> 2a + 24*6 = 136

=> 2a = 136 - 144

=> 2a = -8

=> a = -4.

nth TERM of AP is given by tₙ = a + (n-1)d

31st term is t₃₁ = -4 + (31 - 1)6 = - 4 + 180 = 176.

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The sum of the first 25 terms of an A.P. is 1700, while its common difference is 6. What is the 31st term of the10.A.P.?

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