Question

Find the first four terms of the arithmetic sequence in which the thirteenth term is 97 and the fifieth term is 393

Answer 1

Let the first term of the given AP be a ,

      and the common difference be d

13th term = 97

a + ( 13 - 1 )d = 97

a + 12d = 97

a = 97 - 12d       ------: ( 1 )

50th term = 393

a + ( 50 - 1 ) d = 393

a + 49d = 393

a = 393 - 49d         -----: ( 2 )

      Comparing the values of a from ( 1 ) &  ( 2 )

97 - 12d = 393 - 49d

49d - 12d = 393 - 97

37d = 296

d = 296 / 37

d = 8

          Substituting the value of d in ( 1 )

a = 97 - 12d

a = 97 - 12( 8 )

a = 97 - 96

a = 1

Therefore

                  First term = a = 1

Common Difference = d = 8

Hence,

AP : a , a + d , a + 2d , a + 3d

  ⇒ 1 , 1 + 8 , 1 + 16 , 1 + 24

  ⇒ 1 , 9 , 17 , 25

Therefore first 4 terms of the required AP are 1 , 9 , 17 , 25.

$\:$

Answer 2

Let the first term of the AP be a and the common difference is d.

13th term of AP=97
$a + (n - 1)d = 97 \\ a + (13 - 1)d = 97 \\ a + 12d = 97 \: \: \: ..........(1)$

50th term of the AP=393
$a + (n - 1)d = 393 \\ a + (50 - 1)d = 393 \\ a + 49d = 393 \: \: \: ...........(2)$

eq(2) - eq(1)

$\: \: \: \: a + 49d = 393 \\ - (a + 12d = 97)$
_______________
$37d = 296 \\ d = \frac{296}{37} \\ d = 8$

from eq (1)

$a + 12 \times 8 = 97 \\ a = 97 - 96 \\ a = 1$

so,

$first \: term = a = 1$

$second \: term = a + d = 1 + 8 = 9$

$third \: term = a + 2d = 1 + 2 \times 8 = 17$

$fourth \: term = a + 3d = 1 + 3 \times 8 = 2$

so the first four terms of the given AP are 1 , 9 , 17 and 25.