Question

In an arithmetic progression of 50 terms, the sum of first ten terms is 210 and the sum of last fifteen

terms is 2565. Find the arithmetic progression.​

Answer 1

Answer:

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so here we go

Step-by-step explanation:

here term 1-term 10 out of 50 terms is the set of our first 10 terms

so let the first term be a and that of the tenth term be a+9d

also here term 36-term 50 out of 50 terms is the set of our last 15 terms

so let the 36 term be a+35d and that of the 50th term be a+49d

so now apply formula

Sn=n/2×(t1+tn) in both conditions

so thus according to first condition

sum of first 10 terms=210

so put n=10 in formula t1=a and tn=a+9d

we get

S10=10/2×(a+a+9d)

thus 210=5×(2a+9d)

ie 2a+9d=42 (1)

according to second condition

sum of last 15 terms=2565

again apply same formula

put n=15 t1=a+35d tn=a+49d

thus we get S15=15/2×(a+35d+a+49d)

ie 2565×2=15×(2a+84d)

so 5130/15=2a+84d

ie 2a+84d=342 (2)

subtract (1) from (2)

we get 75d=300

ie d=4

substitute value of d in any of two terms

we get

a=3

so t2=t1+d

=3+4

=7

t3=t2+d

=7+4

=11

t4=t3+d

=11+4

=15

thus the required ap is 3,7,11,15,and so on