If sum of 3rd and 8th terms of an A.P. is 7 and sum of 7th and 14th terms is – 3 then find the 10th term.

Now, By using n^th term of an A.P. formula

t_n = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_n = n^th terms

Hence, by given condition we get,

t₃ + t₈ = 7

⇒ [a + (3 – 1)d] + [a + (8 – 1)d] = 7

⇒ [a + 2d] + [a + 7d] = 7

⇒ 2a + 9d = 7 …..(1)

t₇ + t₁₄ = – 3

⇒ [a + (7 – 1)d] + [a + (14 – 1)d] = – 3

⇒ [a + 6d] + [ a + 13d] = – 3

⇒ 2a + 19d = – 3 …..(2)

Subtracting eq. (1) from eq. (2)

⇒ [2a + 19d] – [2a + 9d] = – 3 – 7

⇒ 10d = – 10

Substituting, “d” in eq. (1)

⇒ 2a + 9 × ( – 1) = 7

⇒ 2a – 9 = 7

⇒ 2a = 7 + 9 = 16

Now, we can find value of t₁₀

Thus, t₁₀ = 8 + (10 – 1)× ( – 1)

⇒ t₁₀ = 8 – 9 = – 1