Question

in an A.P., T1=22,Tn=-11,Sn=66,find n

Answer 1

in an AP ,
$\bf{T_1=22,T_n=-11\:\:S_n=66}$

Let a is the first term and d is the common difference of an AP.
then, $\bf{T_1=22}$
a + (1 - 1) = 22
a = 22 ----(1)

again, $\bf{T_n=-11}$
a + (n - 1)d =-11
22 + (n - 1)d = -11 [ from equation (1),
(n - 1)d = -33 ------(2)

now, $\bf{S_n=66}$
=> n/2[2a + (n - 1)d ] = 66
=> n/2 [ 2 × 22 + (-33)] = 66
[ from equations (1) and (2)]
=> n/2 [ 44 -33 ] =66
=> n/2 × 11 = 66
=> n = 66 × 2/11 = 12

hence, n = 12

Answer 2

While dealing with AP I suggest you to write three formulae before starting to solve:

T(n) = a + (n-1)d

S(n) = n/2 (a+l)

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

In the given problem, we have T(1) = 22, T(n) = -11, S(n) = 66.

Clearly, we have to apply the 2nd formula.

S(n) = n/2 (a+l)

or, 66 = n/2 (22-11)

or, 66 = 11n/2

or, 11n = 132

or, n = 132/11 = 12.

Therefore there are 12 terms.