Question

if second and third term of A.P. are 24 and 28 respectively then find the sum of first 61 terms.​

Answer 1

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.

Answer 2

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: