Question: Numbers $ a _1, $ $ a _2, $ $ a _3, $ $ a _4, $ $ a _5,…, $ $ a _{24} $ with common difference = 10, are in arithmetic progression and $ a _1+a _5+a _{10}+a _{15}+a _{20}+a _{25}=225. $ The value of $ a _1+a _2+a _3+a _4+a _5+…+a _{23}+a _{24} $ is
Options:
A) 525
B) 725
C) 860
D) 900
Show Answer
Answer:
Correct Answer: C
Solution:
- As $ a _1, $ $ a _2, $ $ a _3, $ $ a _4,…., $ $ a _{24} $ are in AP.
So, let $ a _2-a _1=a _3-a _2=a _4-a _3=….=d $
$ \therefore $ $ a _2=a _1+d $
$ \Rightarrow $ $ a _3=a _1+2d $
$ a _4=a _1+3d $
$ \Rightarrow $ $ a _{24}=a _1+23d $
According to the question,
$ a _1+a _5+a _{10}+a _{15}+a _{20}+a _{25}=225 $
$ a _1(a _1+4d)+(a _1+9d)+(a _1+14d) $ $ +(a _1+19d)+(a _1+24d)=225 $
$ \Rightarrow $ $ 6a _1+70d=225 $
(i)
Now, $ a _1+a _2+a _3+a _4+…+a _{24} $
$ =a _1+(a _1+d)+(a _1+2d)+(a _1+3d) $
$ +…+(a _1+23d) $
$ =24a _1+d(1+2+3+4+5+…+23) $
$ =24a _1+d( \frac{23\times 24}{2} ) $
$ =24a _1+276d=4(6a _1+69d) $
$ =4(225-70d+69d) $ [from Eq. (i)]
$ =900-4d $
$ =900-40=860 $ $ [\because d=10,\text{given }] $
BETA
