Question

Smantar sreni T1 = 22, Tn =-11, Sn = 66, n = ?

Answer 1

Given,

          T₁ = 22

          $T_{n}$ = -11

           $S_{n}$ = 66

We know, in AS sum of terms is $\dfrac{n}{2}( T_{1}+T_{n} )$

in the question, all values are given except, so substituting the given values in the formula of sum of n terms.

66 = $\dfrac{n}{2}(22 + ( -11))$

66 = $\dfrac{n}{2}(22-11)$

66 = $\dfrac{n}{2}(11)$

$\dfrac{66\times2}{11} = n$

12 = n

Therefore, n = 12 . It means that there are n 12 terms.

Answer 2

HOLA USER ✌

HERE'S YOUR ANSWER FRIEND,

==> GIVEN : T1 = 22, Tn = -11, Sn = 66, n = ?

WE KNOW,

Sn = n X [T1 + Tn] / 2

==> Sn = n X [22 + (-11)]/2

==> Sn = 66 ............... {given}

THUS,

==> 66 = n X [11/2]

==> 132 = 11n

==> n = 132/11

==> n = 12

IS THE REQUIRED ANSWER.

HOPE IT HELPS YOU FRIEND.

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