Q. 104.0( 20 Votes )

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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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