Question

the s1 term of an ap is 63 the s3 is 72 find s2​

Answer 1

Step-by-step explanation:

s1=63

1/2(2×a+(1-1)d)=63

2a=63×2

2a=126

a=63

then

s3=72

3/2(2a+(3-1)d)=72

2a+2d=72×2/3

2×63+2d=24×2

126+2d=48

2d=48-126

2d= -78

d= -39

then,

s2=2/2(2a+(2-1)d)

1(2×63+1×(-39))

126-39

97

Answer 2

Answer:

s2=63+9/2

Step-by-step explanation:

s1term=63

a=63........(1) eq

s3 term = 72

a+2d=72..........(2)eq

solve eq (2) in eq (1)

a+2d=72

a =63

sub - - -

2d=9 (because + a & - a canceled)

d=9/2

verification

s3 term = 72

a+2d =72

63 +2×9/2 =72

63 + 9 = 72 ( because 2 & 2 get canceled)

s2 term

a+ d

63+9/2

=67.5

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