Question

the 10th term of an ap is 29 and the sum of its first 20 terms is 610 find the sum of its first 30 terms

Answer 1

Answer:

Step-by-step explanation: T10=a+9d=29. -(I)

S20=610=10( 2a +19d)

=> 61=2a +19d. -(ii)

Multiplying 2 to (I),

58=2a+18d. -(iii)

Equating (ii) and (iii),

3=d

Putting in (I),

a +3×9=29

=>a =2

S30=15(4+29×3)

=> S30=15(91)

=> S30=1365

Answer 2

Answer:

Step-by-step explanation:

Its 10the term is 29

a10 = 29

an=a+(n-1)d

29=a + (10-1)d

29=a+9d ---- eqn 1

Given that

S20 = 610

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

610 =20/2(2a+(20-1)d

610 = 10(2a+19d)

61 =2a+19d ---- eqn 2

Multiplying eqn 1 by 2 we get

58 =2a+18d -----eqn 3

Solving eqn 2 and 3 we get

d=3 and a=2

S30=30/2(2*2+(30-1)3)

= 1365