Answer :

Let first term = a

Common difference = d

Since, A.P. consist of 37 terms, therefor the middle most term is

Thus, three middle most term are t_{18} = 18^{th} term, t_{19} = 19^{th} term,

t_{20} = 20^{th} term

We use 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

Thus, on substituting all values we get,

Given, t_{18} + t_{19} + t_{20} = 225

⇒ [a + (18 – 1)d] + [a + (19 – 1)d] + [a + (20 – 1)d] = 225

⇒ [a + 17d] + [a + 18d] + [a + 19d] = 225

⇒ 3a + 54d = 225

Dividing by 3

⇒ a + 18d = 75 …….(1)

Given, sum of last three term is 429

⇒ t_{35} + t_{36} + t_{37} = 429

⇒ [a + (35 – 1)d] + [a + (36 – 1)d] + [a + (37 – 1)d] = 429

⇒ [a + 34d] + [a + 35d] + [a + 36d] = 429

⇒ 3a + 105d = 429

Dividing by 3

⇒ a + 35d = 143 …….(2)

Subtracting eq. (1) from eq. (2) we get,

⇒ [a + 35d] – [a + 18d] = 143 – 75

⇒ 17d = 68

Substituting value of ‘d’ in eq. (1) we get,

⇒ a + 18 × 4 = 75

⇒ a + 72 = 75

⇒ a = 75 – 72 = 3

⇒ a = t_{1} = 3

We know that, t_{n + 1} = t_{n} + d

t_{2} = t_{1} + d = 3 + 4 = 7

t_{3} = t_{2} + d = 7 + 4 = 11

t_{4} = t_{3} + d = 11 + 4 = 15

t_{37} = 3 + (37 – 1) × 4

t_{37} = 3 + 36 × 4

t_{37} = 3 + 144 = 147

Thus, the A.P. is 3, 7, 11, . . . . ., 147

Rate this question :

Attempt any threeMaharashtra Board - Algebra Papers

Attempt any four Maharashtra Board - Algebra Papers

<span lang="EN-USMaharashtra Board - Algebra Papers

<span lang="EN-USMaharashtra Board - Algebra Papers

<span lang="EN-USMaharashtra Board - Algebra Papers

Attempt any threeMaharashtra Board - Algebra Papers

<span lang="EN-USMaharashtra Board - Algebra Papers

<span lang="EN-USMaharashtra Board - Algebra Papers

Attempt any threeMaharashtra Board - Algebra Papers

Attempt any four Maharashtra Board - Algebra Papers