Arithmetic progression
Find the sum No. 4
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Answers
Answered by
5
Hello,
Solution:-) 420
Process:-) This is consisting of two AP's.
S
1
=5+9+13+....+77+81, where a=5,
d=4
S
2
=−41−39−37−...(−3), where a=
−41 and d=2
for
S
1
,81=5+(n−1)×4⇒n−1=
76
4
,
hence n=20
S
20
=
20
2
×[5+81]=10×86=860
for
S
2
,−3=−41+(n−1)×2⇒n−1
=
38
2
,hence n=20
S
20
=
20
2
×[−41−3]=−10×44=−440
Total sum=
S
1
+
S
2
=860−440=420
Answer=420
Solution:-) 420
Process:-) This is consisting of two AP's.
S
1
=5+9+13+....+77+81, where a=5,
d=4
S
2
=−41−39−37−...(−3), where a=
−41 and d=2
for
S
1
,81=5+(n−1)×4⇒n−1=
76
4
,
hence n=20
S
20
=
20
2
×[5+81]=10×86=860
for
S
2
,−3=−41+(n−1)×2⇒n−1
=
38
2
,hence n=20
S
20
=
20
2
×[−41−3]=−10×44=−440
Total sum=
S
1
+
S
2
=860−440=420
Answer=420
Anonymous:
and the negative numbers?
Ur welcome
Answered by
8
Add the positive number with the negative number next to it.
Now you get an AP with common difference of 6 and first term as -36
Tn = a+(n-1)d
78= -36+6n-6
n=120/6 =20
Sum = n/2 * 2a+(n-1)d
= 10* 42
= 420
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