The 5th term & 12th term of anArithematic progression are 30 & 65 resp

1.	The 5th term & 12th term of anArithematic progression are 30 & 65 respectively.Find the sum of its 26th term .
Answer» ANSWER: given, a5=30 a12=65 s26=? a5=a+(n-1)d 30=a+(5-1)d 30=a+4d 30-4d=a --------(i) a12=a+(n-1)d substituting the value of a from (i), 65=30-4d+(12-1)d 65=30-4d+11d 65=30+7d 65-30=7d 35/7=d 5=d substituting the value of d in (i) 30-4(5)=a 30-20=a 10=a s26=n/2[2A+(n-1)d] =26/2[2×10+(26-1)5] =13(20+125) =13×145 =1885

The 5th term & 12th term of anArithematic progression are 30 & 65 respectively.Find the sum of its 26th term .

Answer»

ANSWER:

given,

a5=30

a12=65

s26=?

a5=a+(n-1)d

30=a+(5-1)d

30=a+4d

30-4d=a --------(i)

a12=a+(n-1)d

substituting the value of a from (i),

65=30-4d+(12-1)d

65=30-4d+11d

65=30+7d

65-30=7d

35/7=d

5=d

substituting the value of d in (i)

30-4(5)=a

30-20=a

10=a

s26=n/2[2A+(n-1)d]

=26/2[2×10+(26-1)5]

=13(20+125)

=13×145

=1885

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The 5th term & 12th term of anArithematic progression are 30 & 65 respectively.Find the sum of its 26th term .​

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The 5th term & 12th term of anArithematic progression are 30 & 65 respectively.Find the sum of its 26th term .