if the 5th and 12th terms of an A.P are -4 and -8 respectively, find the sum of first 20 terms of an A.P.
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2
à+4d= -4---------(i)
a+11d= -8---------(ii)
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-7d= 4
d= -4/7
substituting in equation ( I )
a+4d= -4
a+4*(-4/7)=-4
a-16/7= - 4
a= -4+16/7
a= -12/7
sum= 20/2[2* -12/7 +(20-1)-4/7]
s= 10[ -24/7 -76/7]
s= 10[ -100/7]
sum= -1000/7
is this the answer??? ( i am not sure actually ) ...
a+11d= -8---------(ii)
----------------
-7d= 4
d= -4/7
substituting in equation ( I )
a+4d= -4
a+4*(-4/7)=-4
a-16/7= - 4
a= -4+16/7
a= -12/7
sum= 20/2[2* -12/7 +(20-1)-4/7]
s= 10[ -24/7 -76/7]
s= 10[ -100/7]
sum= -1000/7
is this the answer??? ( i am not sure actually ) ...
Akanksha1710:
ok ....
i guess now my amswer is right do check if this is the answer in YOUR worksheet
thank u for marking as the brainliest
Answered by
1
a5=(-4)
a12=(-8)
d = ((-8)-(-4))/7
= (-4/7)
Therefore, a = (-4)-(-4/7)4 = (-12/7)
Now, s20 = (20/2)(2(-12/7)+(19)(-4/7)
= (-1000/7)
a12=(-8)
d = ((-8)-(-4))/7
= (-4/7)
Therefore, a = (-4)-(-4/7)4 = (-12/7)
Now, s20 = (20/2)(2(-12/7)+(19)(-4/7)
= (-1000/7)
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