In an A.P. the 10th term is 46, sum of the 5th and 7th term is 52. Find the A.P.

Given: t₁₀ = 46

t₅ + t₇ = 52

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,

⇒ t₁₀ = 46

⇒ a + (10 – 1)d = 46

⇒ a + 9d = 46 ……(1)

⇒ t₅ + t₇ = 52

⇒ [a + (5 – 1)d] + [a + (7 – 1)d] = 52

⇒ [a + 4d] + [a + 6d] = 52

⇒ 2a + 10d = 52 ……(2)

Multiply eq. (2) by 2 we get,

⇒ 2a + 18d = 92 ……(3)

Subtract eq. (2) by eq. (3)

⇒ [2a + 18d] – [ 2a + 10d] = 92 – 52

⇒ 8d = 40

Substitute “d” in (1)

⇒ a + 9 × 5 = 46

⇒ a + 45 = 46

⇒ a = t₁ = 46 – 45 = 1

we know that, t_{n + 1} = t_n + d

⇒ t₂ = t₁ + d = 1 + 5 = 6

⇒ t₃ = t₂ + d = 6 + 5 = 11

Hence, an A.P. is 1, 6, 11, . . .