The 16th and 17th term of an A.P. are 19 and 41. Find 40th term
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Given, the 6th and 17th term of an A.P. are 19 and 41 respectively.
⇒ a6 = a + (6 -1)d = 19
⇒ a + 5d = 19 ... (1)
Also, a17 = a + (17 -1)d = 41
⇒ a + 16d = 41 ... (2)
On subtracting (1) from (2), we get
a + 16d - (a + 5d) = 41 - 19
⇒ 11d = 22
⇒ d = 2
On putting d = 2 in (1), we get
a + 5(2) = 19
⇒ a = 9
Now, 40th term of the a.p. is:
a40 = a + (40 -1)d = a + 39d
= 9 + 39(2) = 9 + 78 = 87
⇒ a6 = a + (6 -1)d = 19
⇒ a + 5d = 19 ... (1)
Also, a17 = a + (17 -1)d = 41
⇒ a + 16d = 41 ... (2)
On subtracting (1) from (2), we get
a + 16d - (a + 5d) = 41 - 19
⇒ 11d = 22
⇒ d = 2
On putting d = 2 in (1), we get
a + 5(2) = 19
⇒ a = 9
Now, 40th term of the a.p. is:
a40 = a + (40 -1)d = a + 39d
= 9 + 39(2) = 9 + 78 = 87
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