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An A.P. consists of 37 terms. The sum of the three middlemost terms is 225 and the sum of the last three terms is 429. Find the A.P. and the sum of all the 37 terms.
Answers
ANSWER:
- Sum of 37 terms = 2775.
- Required A.P : 3,7,11,15,19 .........
GIVEN:
- Number of terms = 37.
- The sum of the three middle most terms is 225.
- The sum of the last three terms is 429.
TO FIND:
- Sum of all 37 terms and the A.P
SOLUTION
FINDING THE MID-MOST Term:
n is odd so middle term = 37+1/2
= 38/2
= 19
So three mid most terms are :
18th , 19th , 20th term.
Formula :
nth term = a+(n-1)d
Where:
=> a = First term.
=> n = number of terms.
=> d = common difference.
=> 18th term +19th term + 20th term = 225 ....(i)
Now finding 18th ,19th and 20th term and putting in eq(i).
=> [a+(18-1)d]+[a+(19-1)d]+[a+(20-1)d] = 225
=> a+17d+a+18d+a+19d = 225
=> 3a+54d = 225
=> 3(a+18d) = 225
=> a+18d = 225/3
=> a+18d = 75. .....(ii)
Now sum of last three terms = 429.
Last three terms are 35th , 36th and 37th term.
=> 35th term +36th term +37th term = 429
=> [a+(35-1)d]+[a+(36-1)d]+[a+(37-1)d] = 429
=> a+34d+a+35d+a+36d = 429
=> 3a+105d = 429
=> 3(a+35d) = 429
=> a+35d = 429/3
=> a+35d = 143 .,....(iii)
Subtracting eq(ii) from (iii) we get;
=> a+35d-a-18d = 143-75
=> 17d = 68
=> d = 68/17
=> d = 4
Putting d= 4 in eq(ii) we get;
=> a+18(4) = 75
=> a+72 = 75
=> a = 75-72
=> a = 3
Here :
First term(a) = 3
Common difference (d) = 4
Required A.P = 3,7,11,15,19 .........
Formula:
Sum of n terms = n(2a+(n-1)d)/2
Where:
=> a = First term.
=> n = number of terms.
=> d = common difference.
Sum of 37 terms :
= 37(2*3+(37-1)4)/2
= 37(6+144)/2
= (37*150)/2
= 37*75
= 2775