find the sum of first 17 term of ap whose nth term is given by Tn=7-4n
Answers
Answered by
4
T1= 7-4(1)=3
T2=7-4(2)=-1
T3=7-4(3)=-5
T17=7-4(17)=-61
Sn=n/2(a+l)
=17/2(3-61)
=17/2×-58
= -493
T2=7-4(2)=-1
T3=7-4(3)=-5
T17=7-4(17)=-61
Sn=n/2(a+l)
=17/2(3-61)
=17/2×-58
= -493
Answered by
3
GIVEN
( last term) an = 7 - 4n
n (number of terms) = 17
Therefore, (last term) an is achieved
by placing 17 at n
a17 = 7 - (4 x 17)
a17 = 7 - 68
a17 = - 61
Therefore, a (first term) is achieved
by placing 1 at n
= 7 - (4 x 1)
= 7 - 4
= 3
FORMULA SUM
Sn = n (a + an) / 2
Sn = [17 [3 + (- 61)] / 2
Sn = [17(3 - 61)] / 2
Sn = [17 x ( - 58)] / 2
Sn = - (17 x 29)
Sn = - 493
Therefore, Sum of this A. P is - 493.
( last term) an = 7 - 4n
n (number of terms) = 17
Therefore, (last term) an is achieved
by placing 17 at n
a17 = 7 - (4 x 17)
a17 = 7 - 68
a17 = - 61
Therefore, a (first term) is achieved
by placing 1 at n
= 7 - (4 x 1)
= 7 - 4
= 3
FORMULA SUM
Sn = n (a + an) / 2
Sn = [17 [3 + (- 61)] / 2
Sn = [17(3 - 61)] / 2
Sn = [17 x ( - 58)] / 2
Sn = - (17 x 29)
Sn = - 493
Therefore, Sum of this A. P is - 493.
Similar questions