the sum of first 14terms of an AP is 1505 and its first term is 10 .find 14th term
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Answered by
4
Sn = n/2 (2a+(n-1)d
SE = 7(20+13d)
1505/7 = 20+13d
Hence, d=15
Now
14th term = a + (n-1)d
= 10+13×15
= 205 ans.
hOpe this Help.
SE = 7(20+13d)
1505/7 = 20+13d
Hence, d=15
Now
14th term = a + (n-1)d
= 10+13×15
= 205 ans.
hOpe this Help.
Answered by
6
HI !
Given,
a = 10
S₁₄ = 1505
S₁₄ = 14/2(2*10 + 13d)
1505 = 7(20 + 13d)
140 + 91d = 1505
91 d = 1505 - 140
= 1365
d = 1365/91
d = 15
14th term = a₁₄ = a + 13d
10 + 13*15 = 10 + 195 = 205
14th term = 205
Given,
a = 10
S₁₄ = 1505
S₁₄ = 14/2(2*10 + 13d)
1505 = 7(20 + 13d)
140 + 91d = 1505
91 d = 1505 - 140
= 1365
d = 1365/91
d = 15
14th term = a₁₄ = a + 13d
10 + 13*15 = 10 + 195 = 205
14th term = 205
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