Class 10 Math All topics Arithmetic Progressions

If $a_{1}, a_{2}, a_{3},....$ is an A.P. such that $a_{1}+a_{5}+a_{10}+a_{15}+a_{20}+a_{24}=225$ then $a_{1}+a_{2}+a_{3}+...+a_{23}+a_{24}$ is equal to-

(a)

$909$

(b)

$75$

(c)

$750$

(d)

$900$

Correct answer: (d)

Solution: Since $a_{1}+a_{5}+a_{10}+a_{15}+a_{20}+a_{24}=225$

$⇒6a+69d=225$

$⇒2a+23d=75$

Now $S_{24}=a_{1}+a_{2}+a_{3}+...+a_{23}+a_{24}$

$S_{24}=12(2a+23d)$

$=900$

Here we use the following formula to calculate the sum of first n terms of an AP:

$S_{n}=2n (2a+(n−1)d)$

$⇒6a+69d=225$

$⇒2a+23d=75$

Now $S_{24}=a_{1}+a_{2}+a_{3}+...+a_{23}+a_{24}$

$S_{24}=12(2a+23d)$

$=900$

Similar topics

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

surface areas and volumes

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

surface areas and volumes

Related Questions

Related Questions