Question

In an Arithmetic Progression
(A.P.) the fourth and sixth
terms are 8 and 14
respectively, find the Sum of
the first 24 terms *​

Answer 1

Answer:

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Step-by-step explanation:

t4 = a + 3d = 8

t6 = a + 5d = 14

t6 - t5 = 6

t6 - t5 = 2d

2d = 6

d = 3

SUB d = 3 in a +3d = 8

a + 9 = 8

a = -1

S24 = 24/2*[-2+(23)*3] = 12*[67] = 804

Answer 2

Answer:

Sn=804

Step-by-step explanation:

a+5d=14---(1)

a+3d=8-----(2)

a=8-3d

putting this in eq 1

8-3d+5d=14

2d=6

d=3

a=8-3(3)

a= -1

Sn=n/2(2a+(n-1)d)

Sn=24/2(2(-1)+(24-1)3)

Sn=12(-2+69)

Sn= -24+828

Sn=804

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