Question

The seventh term of an A.P is 15 and the fourth term is 9.find the first term and common difference

Answer 1

Let a be the first term and d be the common difference of the A.P.

$Now, 4th  \: term \: of \: the \: A.P. = a+3d=22 ... (1)$

$Also,  \: 15th  \: term \: of \: the \: A.P. = a+14d=66 ... (2)$

Subtracting (1) from (2), we get

$11d=44⟹d=4$

Substituting value of d in (1), we have

$a+12=22⟹a=10$

$Sum \: of \: first \: 8 \: terms =  \frac{n}{2} (2a+(n−1)d)$

$= > \frac{8}{2} (20+(8−1)4)=192$

Hence, the first term, the common difference and the sum upto 8 terms are 10,4 and 192 respectively.

Answer 2

Answer:

According to the formula of A.P.:

$\tt \: a_7 = a + (7 - 1)d \\ \\ \tt \implies a + 6d = 15 \: \: \: \: \: - - - - (i)$

$\tt \: a_4 = a + (4 - 1)d \\ \\ \tt \implies \: a + 3d = 9 \: \: \: \: - - - - - (ii)$

Subtracting (I) and (ii)

$\tt \: 3d=15-9 \\ \\ \tt \: 3d = 6 \\ \\ \tt \: d = \cancel\frac{6}{3} \\ \\ \tt \: d = 2$

d = 2 is the common difference.

Putting the value of d=2 in (I)

$\tt \: a+6d=15 \\ \\ \tt \: a + 6 \times 2 = 15 \\ \\ \tt \: a + 12 = 15 \\ \\ \tt \: a = 15 - 12 \\ \\ \tt \: a = 3$

First term of this AP is 3 and common difference is 2.