7 ,7+r, *, *, *, *, *, *, *, 14+2r,
Term n = 7 +r(n-1)
term 10 = 7 +9 r = 14 + 2 r
so 7 r = 7
r = 1
and we have
7 ,8, 9, 10, 11, 12, 13, 14, 15, 16,
n = 19
term 19 = 7 + 1(18) =25
Given that the first term of an A.P is 7 and its 10th term is twice its second calculate a)19th term b) sum of 28 term c) different between the 9th and 6th term
6 answers
1st term = 7, 2nd term = a+d = 7+d. Then, since 10th term is twice it second, it implies 10th term = a+9d = 7+9d = 2(7+d). Open the bracket and collect like terms we have, 9d-2d = 14-7. Then 7d = 7. Finally, d = 1. (a) the 19th terms = a+18d = 7+18(1) = 25. (b) the sum of 28th term = n/2[2a + (n-1)d] = 28/2[2(7) + (27(1)] = 574. Hence, to solve for (c), find the 9th term and the 6th term separately, then take the difference of the answers. You will have 15-12 = 3.
a)T19=12
b)S28=48
b)S28=48
Good And Happy
Once
a=7
a+9d=2(a+d)
7+9d=2(7+d)
7+9d=14+2d
14-7=9d-2d
7÷7=7d÷7
d=1
1. a+18d
7+18(1)
=25
2. Sn=n÷2[2a+(n-1)d]
28÷2[2(7)+(28-1)1]
14[14+27]
14[41]
=574
3. 9th term=a+8d
7+8(1)
=15
6th term=a+5d
7+5(1)
=12
Difference=15-12=3
a+9d=2(a+d)
7+9d=2(7+d)
7+9d=14+2d
14-7=9d-2d
7÷7=7d÷7
d=1
1. a+18d
7+18(1)
=25
2. Sn=n÷2[2a+(n-1)d]
28÷2[2(7)+(28-1)1]
14[14+27]
14[41]
=574
3. 9th term=a+8d
7+8(1)
=15
6th term=a+5d
7+5(1)
=12
Difference=15-12=3