Question

Find the sum of 30 terms of an AP in which 11TH term is 0 and 10th term is one more than twice the 5th term

Answer 1

Answer:

135/11

Step-by-step explanation:

a = First term of A.P.

d = Common difference of A.P.

To find, nth term of A.P.

=> nth term = a + ( n - 1 )d

For solution, Please refer to the attachment....

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Answer 2

The sum of the first 30 terms is 135/11.

Step-by-step explanation:

Let the first term of the AP is a and the common difference is d.

∴ t₁₁ = a + (11 - 1) d = a + 10d,

t₁₀ = a + (10 - 1) d = a + 9d &

t₅ = a + (5 - 1) d = a + 4d

By the given conditions,

t₁₁ = 0

or, a + 10d = 0 ..... (1)

& t₁₀ = 2 t₅ + 1

or, a + 9d = 2 (a + 4d) + 1

or, a + 9d = 2a + 8d + 1

or, a - d + 1 = 0

or, - 10d - d + 1 = 0

or, 11d = 1

or, d = 1/11

∴ a = - 10d, by (1)

= - 10/11

Therefore, the sum of the first 30 terms is

S₃₀ = 30/2 * [ 2 (- 10/11) + (30 - 1) * 1/11 ]

= 15 * [ - 20/11 + 29/11 ]

= 15 * ( 9/11 )

= 135/11

Related question :

In an AP, 1 + 4 + 7 + 10 + ... + X = 287. Find the value of X. - https://brainly.in/question/15720003