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Question

The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.

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Solution

It is given that,
t4 = –15
t9 = –30

Now,
tn=a+n-1dt4=a+4-1d-15=a+3da=-15-3d ...1t9=a+9-1d-30=a+8da+8d=-30-15-3d+8d=-30 from1-15+5d=-305d=-30+155d=-15d=-3a=-15-3-3 from1a=-15+9a=-6

Hence, the given A.P. is –6, –9, –12, ....

Now,
Sn=n22a+n-1dS10=1022a+10-1d =52-6+9-3 =5-12-27 =5-39 =-195

Hence, the sum of the first 10 numbers is –195.

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