Q. 104.3( 16 Votes )

There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.

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 t18 = 18th term, t19 = 19th term,


t20 = 20th term


We use nth term of an A.P. formula


tn = a + (n – 1)d


where n = no. of terms


a = first term


d = common difference


tn = nth terms


Thus, on substituting all values we get,


Given, t18 + t19 + t20 = 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


t35 + t36 + t37 = 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 = t1 = 3


We know that, tn + 1 = tn + d


t2 = t1 + d = 3 + 4 = 7


t3 = t2 + d = 7 + 4 = 11


t4 = t3 + d = 11 + 4 = 15


t37 = 3 + (37 – 1) × 4


t37 = 3 + 36 × 4


t37 = 3 + 144 = 147


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


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