the sum of the 6 terms of AP is 345 and the difference between first term and last term is 55 find the AP
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Let the six terms be a a+d a+2d a+3d a+4d a+5d ------ (*)
Given, a+a+d+a+2d+a+3d+a+4d+a+5d = 345
6a + 15d = 345 ------------------ (1)
Given, a + 5d - a = 55
5d = 55
d = 11. ------------------- (2)
Substitute equation (2) in (1), we get
6a + 15(11) = 345
6a = 345 - 165
6a = 180
a = 30 ------------------ (3)
On Substituting d = 11 and a = 30 in (*) we get the AP values as
30,41,52,63,74,85.
Hope this helps!
Given, a+a+d+a+2d+a+3d+a+4d+a+5d = 345
6a + 15d = 345 ------------------ (1)
Given, a + 5d - a = 55
5d = 55
d = 11. ------------------- (2)
Substitute equation (2) in (1), we get
6a + 15(11) = 345
6a = 345 - 165
6a = 180
a = 30 ------------------ (3)
On Substituting d = 11 and a = 30 in (*) we get the AP values as
30,41,52,63,74,85.
Hope this helps!
Hemamalini15:
Thanks
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