Question 1024637
The nth term of an arithmetic progression is defined as
:
Xn = a + d(n-1), where a is the first term and n is the nth term number
:
We are given two equations
:
23 = a + d(49-1)
37 = a + d(62-1)
:
23 = a +48d
37 = a +61d
:
solve first equation for a
a = 23 - 48d
:
now substitute for a in second equation
:
37 = (23 -48d) +61d
13d = 14
d = 14/13
:
substitute for d in first given equation
:
23 = a + (14/13) * (48)
23 = a + 51.692307692
a = −28.692307692 approx -28.69
:
:
we want the sum of the first 110 terms, the sum of the first n terms of an  arithmetic sequence is
:
Sn = (n/2)(2a + (n-1)d)
:
S110 = (110/2)(2(-28.69) + (109)(14/13))
S110 = (55)((−57.38) + 117.38)
S110 = (55)(60)
:
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S110 = 3300,   answer is option 2
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