if second and third term of A.P. are 24 and 28 respectively then find the sum of first 61 terms.
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Answered by
9
A2 = 24
a + d = 24 – ( 1 )
A3 = 28
a + 2d = 28 – ( 2 )
( 2 ) – ( 1 )
a + 2d – a – d = 28 – 24
d = 4
put in (1)
a+ 4 = 24
a = 24 – 4
a = 20
Hence,
Sn = n/2 [ 2a + ( n – 1 ) d ]
= 61/2 [ 2 (20) + ( 60 ) ( 4 ) ]
= 61 / 2 [ 40 + 240 ]
= 61 / 2 [ 280 ]
= 61 × 140
= 8540 is the answer.
Answered by
3
Answer:
t2 = 24
t3 = 28
d = t2 -t3
d = 4
ti(a) = t2-d
a = 20
to find : sum of first 61 terms
t61 = a+ (n-1)d
t61 = 20 +( 61-1)4
t61= 20 + 240
t61(L) = 260
Sn = (n/2)(a+ l )
Sn = (61 / 2 ) X (20 +260 )
Sn = (61 X 280 ) / 2
Sn = 61 X 140
Sn = 8540 .00
THE SUM OS FIRST 61 TERMS OF a.p IS 8540
Step-by-step explanation:
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