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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.

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Sol.  Let the first term and the common difference of the A.P are a and d respectively.
Since the A.P contains 37 terms, so the middle most term is

$({{37+1}\over2})$th  term = 19th term
Thus three middle most term of this
A.P are 18th, 19th and 20th terms.
t18 + t19 + t20 = 225
a + 17d + a + 18d + a + 19d = 225

[ tn = a + (n –1)d]
3a + 54d = 225

(Dividing both sides by 3 throughout)
According to condition (II);
a + 18d = 75                                 ...(I)
Now,  t35 + t36 + t37 = 429
a + 34d + a + 35d + a + 36d = 429
[tn = a + (n –1)d]
3a + 105d = 429
(Dividing both sides by 3 throughtout)
a + 35d = 143                                  ...(II)

Subtracting (I) from (II)
a + 35d = 143
–         a + 18d =   75

(–)    (–)      (–)

17d =   68
d = 4
Substituting d = 4, in equation (I)
a + 18 × 4 = 75
a + 72 = 75
a = 75 – 72
a = 3
∴      t1 = a = 3
t2 = t1 + d = 3 + 4 = 7
t3 = t2 + d = 7 + 4 = 11

∴      The Required AP is 3, 7, 11, ...

Arithmetic Progression September 28 , 2018 0 Comments 343 views