Help me with this A.P. question please!!

The sum of the first four terms of an arithmetic progression is 36 and the sum of the next ten terms is 370. Find the first term and the common difference.

Please show the steps to solve this. Thank you!

Sum S = (n/2)[2a + (n - 1)d ]

n= 4, S = 36

36 = 2[2a + 3d]
18 = 2a + 3d......(1)
...............................

n = 4 + 10 = 14,
S = 36 + 370 = 406

406 = 7[2a + 13d]
58 = 2a + 13d....(2)
................................

Now solve the simultaneous equations (1) and (2) by subtracting (1) from (2)

40 = 10d
d = 4

Substitute into (1)
18 = 2a + 3(4)
6 = 2a
a = 3

If the sum of terms 5 to 14 is 370 then the sum of the first 14 terms is 406.

=> S4 = (4/2)[2a + 3d] and S14 = (14/2)[2a + 13d]....where a is first term and d is the common difference.

=> 2[2a + 3d] = 36 and 7[2a + 13d] = 406

so, 2a + 3d = 18......(1)

and 2a + 13d = 58...(2)

i.e. 10d = 40

=> d = 4 => a = 3

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