Question

The sixth term of an arithmetic progression is 23 and the sum of first ten terma is 200. Find the seventh term.

Answer 1

An = a+(n-1)d
A7 = a+(6-1)d=23
= a+5d=23 *********** 1 St equation
Sn = n/2 [2a+(n-1)d]
S10 = 10/2 [2a+9d] =200
= 5 [2a +9d]=200
= 10a + 45d =200
= a+45d=20 ************2nd equation
Now equating both 1st and 2nd equations
a +5d = 23
a+ 45d= 20
(-) (-) (-)
____________
40d = 3
d = 3/40
Substitute d= 3/40 In 1st equation
a + 5(3/40)=23
a +(3/8)=23
a= 23-3/8
a = 184-3/8
a = 181/3
= 6.3
Then substitute in an formula
I think the question is wrong