Question

how many terms of series -1+3+7.... be added to get 95 as sum

Answer 1

Hiii friend,


AP = -1,3,7......


Here,


First term (A) = -1

Common difference (D) = 4

Sn = N/2 × [2A +(N-1) × D]

95 = N/2 × [ 2 × -1 + (N-1) × 4]

95 = N/2 × ( -2 + 4N -4)

95 = N/2 × -6+4N

95 = N × -3 + 4N


-3N +4N = 95


N = 95

Hence,

95 term must be taken to get the of 95 as their sum.


HOPE IT WILL HELP YOU...... :-)

Answer 2

Heya!! ✌

➡Here's your answer friend,

Here a(first term) : -1

d(common difference) : 3 - (-1) = 4

Sn = 95

==> Sn = n/2 [2a + (n - 1)d]

==> 95 = n/2 [2 (-1) + (n - 1)4]

==> 95 = n/2 [-2 + 4n -4]

==> 190 = n[ -6 + 4n ]

==> 190 = -6n + 4n²

==> 4n² - 6n = 190

==> 2n² - 3n - 95 = 0

==>by quadratic formula,

==> -b +-√b² - 4ac / 2a

==> -(-3) +- √(-3)² - 4(2)(95) / 2(2)

==> 3 +- √769 / 4

==> x = 3 + 27.73 / 4 or x = 3 - 27.73/4

==> x = 3 + 6.9325 or x = 3 - 6.9325

==> x = 9.9325 approx (9) or x = -3.9325 (neglect)

is the required answer.

Hence, 9th term of A.P. will give sum as 95.

⭐ Hope it helps you : ) ⭐