Question

in arithmetic sequence find: S10 , if t6+t10 = −50, S7 = −91

Answer 1

Solution :

Let, the first term of the sequence is a and the common difference is d

Then, t₆ = a + 5d

& t₁₀ = a + 9d

Given, t₆ + t₁₀ = - 50

or, a + 5d + a + 9d = - 50

or, 2a + 14d = - 50

or, a + 7d = - 25 ..... (i)

Also, S₇ = - 91

or, 7/2 * [2a + (7 - 1)d] = - 91

or, 7 (2a + 6d) = - 2 * 91

or, 14a + 42d = - 182

or, a + 3d = - 13 ..... (ii)

Subtracting (ii) from (i), we get

a + 7d - a - 3d = - 25 + 13

or, 4d = - 12

or, d = - 3

From (i), we get

a + 7 (- 3) = - 25

or, a - 21 = - 25

or, a = - 25 + 21

or, a = - 4

We have solved that

first term (a) = - 4

& common difference (d) = - 3

Hence, the required sequence of arithmetic progression be

- 4, - 4 - 3, - 4 - 3 - 3, - 4 - 3 - 3 - 3, ...

i.e., - 4, - 7, - 10, - 13, - 16, - 19, ...

Answer 2

Let, the first term of the sequence is a and the common difference is d

Then, t₆ = a + 5d

& t₁₀ = a + 9d

Given, t₆ + t₁₀ = - 50

or, a + 5d + a + 9d = - 50

or, 2a + 14d = - 50

or, a + 7d = - 25 ..... (i)

Also, S₇ = - 91

or, 7/2 * [2a + (7 - 1)d] = - 91

or, 7 (2a + 6d) = - 2 * 91

or, 14a + 42d = - 182

or, a + 3d = - 13 ..... (ii)

Subtracting (ii) from (i), we get

a + 7d - a - 3d = - 25 + 13

or, 4d = - 12

or, d = - 3

From (i), we get

a + 7 (- 3) = - 25

or, a - 21 = - 25

or, a = - 25 + 21

or, a = - 4

We have solved that

first term (a) = - 4

& common difference (d) = - 3

Hence, the required sequence of arithmetic progression be

- 4, - 4 - 3, - 4 - 3 - 3, - 4 - 3 - 3 - 3, ...

i.e., - 4, - 7, - 10, - 13, - 16, - 19, ...

hope it helps :