If $a_{1}, a_{2}, a_{3}$ ,.are in $A . P .$ 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

909

No worries! We‘ve got your back. Try BYJU‘S free classes today!

75

No worries! We‘ve got your back. Try BYJU‘S free classes today!

750

No worries! We‘ve got your back. Try BYJU‘S free classes today!

900

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $900$
b'
Given, $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$
We have $a_{1} = a$ , $a_{5} = a + 4 d$ ,... $a_{24} = a + 23 d = 225$
Sum of this, $6 a + 69 d = 225$
$a_{1} + a_{2} + a_{3} + a_{4} + . . . . . . . . . . . . . + a_{23} + a_{24}$
$a_{1} = a$ , $a_{2} = a + d$ ..........., $a_{24} = a + 23 d = 225$
Sum of this is $= 24 a + 306 d = 4 (6 a + 69 d) = 4 \times 225 = 900$
'

Suggest Corrections

Similar questions

Q. If

a_{1}, a_{2}, a_{3}

,.... are in 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} =

If $a_{1}, a_{2}, a_{3} . . . . .$ are in $A . P .$ such that $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225$ .

Find

a_{1} + a_{2} + a_{3} . . . . . . . . . + a_{23} + a_{24}

= ?

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

Q. If

a_{1}, a_{2}, a_{3}, . . . .

are in 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_{24}

is equal to

Q. If

a_{1}, a_{2}, a_{3}, \dots

are 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} + \dots a_{23} + a_{24}

is equal to

Join BYJU'S Learning Program

If a1,a2,a3,.are in A.P. such that a1+a5+a10+a15+a20+a24=225, then a1+a2+a3+...+a23+a24 is equal to

The correct option is B 900b'Given,a1+a5+a10+a15+a20+a24=225We have a1=a, a5=a+4d,...a24=a+23d=225Sum of this,6a+69d=225a1+a2+a3+a4+.............+a23+a24a1=a,a2=a+d...........,a24=a+23d=225Sum of this is=24a+306d=4(6a+69d)=4×225=900'

If $a_{1}, a_{2}, a_{3}$ ,.are in $A . P .$ 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