The 7th term of an A.P.3,10,17,... is 32 and its 13th term is 62. Find the A.P.
Answers
Given: The 7th term of an A.P is 32 and its 13th term is 62.
Solution :
A. P. is 2, 7, 12, 17, 22, 27, 32, .....
We have , a7 = 32, a13 = 62
Let 'a' be the first term and common difference be 'd'.
nth term , an = a + (n -1)d
a7 = a + (7 - 1)d
32 = a + 6d …………..(1)
a13 = a + (13 - 1)d
62 = a + 12d…………..(2)
On subtracting equation (1) from (2)
(a + 12d) - (a + 6d) = 62 - 32
a + 12d - a - 6d = 30
a - a + 12d - 6d = 30
6d = 30
d = 30/6
d = 5
On substituting the value of d = 5 in eq 1,
32 = a + 6d
32 = a + 6 × 5
32 = a + 30
a = 32 - 30
a = 2
First term , a = 2
Second term , a2 = (a + d) = 2 + 5 = 7
Third term , a3 = (a + 2d) = 2 + 2 × 5 = 2 + 10 = 12
Fourth term , a4 = (a + 3d) = 2 + 3 × 5
a4 = 2 + 15 = 17
Fifth term ,a5 = (a + 4d) = 2 + 4 × 5 = 2 + 20 = 22
Hence , A. P. is 2, 7, 12, 17, 22, 27, 32, .....
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Answer: 2, 7, 12, 17, 22, 27, 32 .....
Given AP: 3,10,17,..
a7 = 32, a13 = 62
Let 'a' be the first term and common difference be 'd'.
As, an = a + (n -1)d
a7 = a + (7 - 1)d
32 = a + 6d ___(1)
a13 = a + (13 - 1)d
62 = a + 12d ___(2)
Subtracting equation (1) from (2)
(a + 12d) - (a + 6d) = 62 - 32
a + 12d - a - 6d = 30
6d = 30
d = 5
Substituting the value of d, in eq 1:
32 = a + 6d
32 = a + 6 × 5
32 = a + 30
a = 2
First term = a = 2
Second term = a2 = (a + d) = 2 + 5 = 7
Third term = a3 = (a + 2d) = 2 + 2 × 5 = 2 + 10 = 12
Fourth term = a4 = (a + 3d) = 2 + 3 × 5 = 2 + 15 = 17
Fifth term = a5 = (a + 4d) = 2 + 4 × 5 = 2 + 20 = 22
Thus, A. P. is : 2, 7, 12, 17, 22, 27, 32 .....