The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
Answers
Answer:
The 40th term is 87.
Step-by-step explanation:
Given :
6th term (a6) = 19
17th term (a17) = 41
an = a + (n – 1) d
a6 = a + (n – 1) d
19 = a + (6 - 1)d
19 = a + 5d ……….(1)
a17 = a + (n – 1) d
41 = a + (17 - 1) d
41 = a + 16d ………….(2)
On Subtracting eq. (1) from eq. (2),
(a + 16d) -( a + 5d) = 41 - 19
a + 16d - a - 5d = 22
16d - 5d = 22
11d = 22
d = 22/11
d = 2
On substituting the value of d = 2 in (1),
19 = a + 5d
19 = a + 5(2)
19 = a + 10
a = 19 - 10
a = 9
40th term :
an = a + (n – 1) d
a40 = 9 + (40 - 1) 2
[a = 9 , d = 2, n = 40]
a40 = 9 + 39 × 2
a40 = 9 + 78
a40 = 87
40th term = 87
Hence, the 40th term is 87.
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Given :
6th term of an AP = 19
a + 5d = 19 -----(1)
17th term of an AP = 41
a + 16d =41-------(2)
Solve equations - (1) & (2) to find difference (d)
a + 5d = 19
a + 16d = 41
(-)
------------------
-11d = -22
d = 22/11
d = 2
substitute d in any equation to find first term (a)
a + 5(2) = 19
a + 10 = 19
a = 19 - 10
a = 9
Find 40th term :
a = 9 d = 2
40th term = a + 39d
= (9) + 39(2)
= 9 + 78
= 87
Therefore, 40th term is 87