Question

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

Answer 1

a1 + a5 + a10 +a15 + a20 + a24  = 225        (given)

(a1+a24) + (a5+a20) + (a10+a15) = 225            .................1

let first term is a & common difference is d then

 a1 = a       ,  a24 = a+23d   ,       a1+a24 = 2a+23d            ..........2

 

 a5 = a+4d ,  a20 = a+19d ,         a5+a20 = 2a+23d           ..............3

 

 a10 = a+9d , a15 = a+14d ,        a10+a15 = 2a+23d          ..............4

putting 2 , 3 , 4 in eq 1 we get

3(2a+23d) = 225

 2a+23d = 75             ...........5

now , a1 + a2 + a3 ............a24 = S24

   S24 = 24/2[2a+(24-10d)]

          =12(2a+23d)             ...............6

from 5 & 6

S24 = 12*75 = 900

Answer 2

Answer:

The value of $a_1+a_2+a_3+a_4+........+a_{24}$ is 900.

Step-by-step explanation:

Given information:

$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$

Concept used:

Arithmetic sequence:

It is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. It is also called Arithmetic Sequence.

The general term of an AP: $a_n=a+(n-1)d$ where $a_n$ represents the nth term, $a$ represents the first term, $n$ represents the number of terms and $d$ represents the common difference of an AP.

Step 1 of 2:

Here, $a_1=a, a_2=a+d, a_3=a+2d,............ ,a_{24}=a+23d$.

$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225\\a+(a+4d)+(a+9d)+(a+14d)+(a+19d)+(a+23d)=225\\6a+69d=225\\2a+23d=75 ....(1)$

Step 2 of 2:

Now, find the value of $a_1+a_2+a_3+a_4+........+a_{24}$

$a_1+a_2+a_3+a_4+......+a_{24}=a+(a+d)+(a+2d)+(a+3d)+.........+(a+23d)\\=24a+d(1+2+3+4+.....+23)\\=24a+d\times \dfrac{23(23+1)}{2}\\=24a+276d\\=12(2a+23d)\\=12\times 75\\=900$

Hence, the value of $a_1+a_2+a_3+a_4+........+a_{24}$ is 900.