The 4th and 7th term of an AP. are 17and 23repectively find a15
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Given, 4 th and 7th term of an A.P is 17 and 23.
since , we know
an = a + ( n - 1 ) d
a4 = a + ( 4 - 1 ) d = a + 3d.
a7= a + ( 7 - 1 ) d = a + 6d
according to question,
a4 = 17
a + 3d = 17 --- (1)
a7 = 23
a + 6d = 23 -- (2)
subtract .(1) and (2) , we get
a + 3d = 17
a + 6d = 23
_________-
-3d = - 6 => d = 2
put value of 'd' in eq. (1) , we get
a + 3( 2 ) = 17
a = 17 - 6 = 11
therefore,
a15 = a + ( 15 - 1 )d = a + 14d
a15 = 11 + 14 (2 ) = 11 + 28 =39
a15 = 39
Your Answer : a15 = 39
_______________________________
since , we know
an = a + ( n - 1 ) d
a4 = a + ( 4 - 1 ) d = a + 3d.
a7= a + ( 7 - 1 ) d = a + 6d
according to question,
a4 = 17
a + 3d = 17 --- (1)
a7 = 23
a + 6d = 23 -- (2)
subtract .(1) and (2) , we get
a + 3d = 17
a + 6d = 23
_________-
-3d = - 6 => d = 2
put value of 'd' in eq. (1) , we get
a + 3( 2 ) = 17
a = 17 - 6 = 11
therefore,
a15 = a + ( 15 - 1 )d = a + 14d
a15 = 11 + 14 (2 ) = 11 + 28 =39
a15 = 39
Your Answer : a15 = 39
_______________________________
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