Question

the sum of first 11 terms of an A.P. is 429 and that of the first 23 terms is 1863. find the A.P.??

Answer 1

s11 = 429 : This means therefore
s11 = n/2(2a+(n-1)d)
429 = 11/2(2a + 10d)
429 x 2/11 = 2(a+5d)
78/2=a+5d
39 = a +5d FIRST

s23 = 1863
s23 = n/2(2a+(n-1)d)
1863 = 23/2(2a + 22d)
1863 x 2/23= 2(a+11d)
162/2=a+11d
81 = a +11d SECOND

ELIMINATE
81=a+11d
39=a+5d
- - -
42=6d
:d =7

81=a+5(7)
:a=4

:AP = 4,11,18...

Answer 2

Step-by-step explanation:

s11 = 429 : This means therefore

s11 = n/2(2a+(n-1)d)

429 = 11/2(2a + 10d)

429 x 2/11 = 2(a+5d)

78/2=a+5d

39 = a +5d FIRST

s23 = 1863

s23 = n/2(2a+(n-1)d)

1863 = 23/2(2a + 22d)

1863 x 2/23= 2(a+11d)

162/2=a+11d

81 = a +11d SECOND

ELIMINATE

81=a+11d

39=a+5d

- - -

42=6d

:d =7

81=a+5(7)

:a=4

:AP = 4,11,18...

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