Question

If t_n represents n^th term of an A.P., t₂ + t₅ – t₃ = 10 and t₂ + t₉ = 17, find its first term and its common difference.

If \[t_n\] represents \[n^{\text{th}}\] term of an A.P., \[t_2 + t_5 - t_3 = 10\] and \[t_2 + t_9 = 17\], find its first term and its common difference.

If the first term of an A.P. is \[a\] and its common difference is \[d.\]

\[t_2 + t_5 - t_3 = 10\]

⇒ \[(a + d) + (a + 4d) - (a + 2d) = 10\]

i.e. \[a + 3d = 10 \qquad \text{.....I}\]

Also, \[t_2 + t_9 = 17\]

⇒ \[(a + d) + (a + 8d) = 17\]

i.e. \[2a + 9d = 17 \qquad \text{.....II}\]

Solve the two equations to get the values of \[a\] and \[d\]

Answer 1

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

The general term of an A.P. is given by t_n = a + (n – 1)d

Now, t₂ + t₅ – t₃ = 10

`=>` (a + d) + (a + 4d) – (a + 2d) = 10

`=>` a + d + a + 4d – a – 2d = 10

`=>` a + 3d = 10  ...(i)

Also, t₂ + t₉ = 17

`=>` (a + d) + (a + 8d) = 17

`=>` 2a + 9d = 17 ...(ii)

Multiplying equation (i) by 2, we get

2a + 6d = 20  ...(iii)

Subtracting (ii) from (iii), we get

–3d = 3

`=>` d = –1

Substituting value of d in (i), we get

a + 3(–1) = 10

`=>` a – 3 = 10

`=>` a = 13

Hence, a = 13 and d = –1.