Solve the following:-
Answers
Step-by-step explanation:
Let the first term be a and common difference be d
Middle term of 29 terms = n + 1
2
= 29 + 1
2
= 30
2
= 15
The middle most terms are 14th, 15th and 16th term
tn = a + (n - 1) d
According to the first condition,
t14 + t15 + t16 = 375
a + (14 - 1) d + a + (15 - 1) d + a + (16 - 1) d = 375
a + 13d + a + 14d + a + 15d = 375
3a + 42d = 375
Dividing throughout by 3,
a + 14d = 125 ... (i)
According to the second condition,
t27 + t28 + t29 = 531
a + (27 - 1) d + a + (28 - 1) d + a + (29 - 1) d = 531
a + 26d + a + 27d + a + 28d = 531
3a + 81d = 531
Dividing throughout by 3,
a + 27d = 177 ...(ii)
Subtracting equation i and ii,
a + 27d = 177
- a + 14d = 125
13d = 52
d = 52
13
d = 4
Substituting value of d in equation i,
a + 14(4) = 125
a + 56 = 125
a = 125 - 56
a = 69
The AP is
a = 69
a + d = 69 + 4 = 73
a + 2d = 69 + 2(4) = 69 + 8 = 77
a + 3d = 69 + 3(4) = 69 + 12 = 81
a + 4d = 69 + 4(4) = 69 + 16 = 85
The required AP is 69, 73, 77, 81, 85, ...