How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?
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Heya Frnd........
a = -14
a₅ = 2
a₅ = a + 4d
=> 2 = -14 + 4d
∴ d = (2+14)/4 = 4 .............(1)
Now,
Sn = n/2 ( 2a + (n-1) d)
40 = n/2 ( 2(-14) + 4n -4 ) .........{ d = 4}
=> 20 = n ( -28 -4 + 4n)
=> 20 = -33n
∴ n = -20/33 ...........ANS
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# HridayAg0102
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a = -14
a₅ = 2
a₅ = a + 4d
=> 2 = -14 + 4d
∴ d = (2+14)/4 = 4 .............(1)
Now,
Sn = n/2 ( 2a + (n-1) d)
40 = n/2 ( 2(-14) + 4n -4 ) .........{ d = 4}
=> 20 = n ( -28 -4 + 4n)
=> 20 = -33n
∴ n = -20/33 ...........ANS
______________________
HOPE IT WILL HELP U..........
PLZ MARK MY ANSWER AS BRAINLIEST IF U LIKE IT..........
☺☺☺☺☺☺☺☺☺☺☺
______________________
# HridayAg0102
✪ BRAINLY BENEFACTOR ✪
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