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1. ARITHMETIC

PROGRESSIONS

  • Sum based on ‘tn’ and ‘Sn’

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Q.6) Obtain the sum of the 56 terms of an A. P. whose

19th and 38th terms are 52 and 148 respectively.

Given :

t19 = 52

t38 = 148

To find :

S56 = ?

tn =

a + (n – 1) d

We know

Now, t19 =

a + (19 – 1) d

52 =

a + 18d

i.e. a + 18d = 52

……. (1)

Also t38 =

a + (38 – 1)d

∴ 148 =

a + 37d

i.e. a + 37d =

148

……. (2)

Adding (1) and (2)

a + 18 d =

52

a + 37 d =

148

=

2a +

55d

200

……. (3)

Now Sn =

∴ S56 =

=

=

[from (3)]

=

∴ S56 =

∴ Sum of first 56 terms of A.P. is 5600.

For given value of t19

For given value of t38

Lets add the 2

equations

replace n by 19

replace n by 38

Same co–efficient &

Same sign

To find, S56 replace

n by 56 in Sn

 

d = 5.05…

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