Sum of First 24 Terms

$S_{24}$ = $\frac{24}{2} (2a_1 + 23d)$ , where $a_1$ is the first term and d is the common difference.

$S_{24}$ = 24 $a_{1}$ + 276d.

Now, 225 = $a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24}$

225 = $6a_1 + 69d$

We can see that, 24 $a_{1}$ + 276d = $4 \times (6a_1 + 69d)$

So, $S_{24}$ = $4 \times 225$ = $\boxed{900}$

we know that terms in AP are a,a+d,a+2d.....so on then sum of first 24 terms is equal to a+a+d+a+2d+......+a+23d, which is equal to 24a+276d....., Given that a1+a5+a10+a15+a20+a24=225, which is simplified to 6a+69d=225, hence 4*(6a+69d)=24a+276d=900

a {1} = a & a {5} = a +4d & a {10} = a +9d & a {15} = a +14d & a {20} = a +19d & a {24} = a +23d

Given, 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 => 3 (2a + 23d) = 225 So, 2a + 23d = 75

Now, S_{24} = \frac{24}{2} (2a + 23d) = 12 \times 75 = \boxed{900}

a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₄ = 225 $$ => a + [a+(5−1)d] + [a+(10−1)d] + [a+(15−1)d] + [a+(20−1)d] + [a+(24−1)d] = 225 $$ => a + (a + 4d) + (a + 9d) + (a + 14d) + (a + 19d) + (a + 23d) = 225 $$ => 6a + 69d = 225 $$ => 2a + 23d = 75 $$ $$ Now, $$ a₁ + a₂ + a₃ + a₄ + a₅ + .... + a₂₄ $$ = (½) x [a₁ + a₂₄] x 24 $$ = (½) x [a + a +(24−1)d] x 24 $$ = (½) x [a + a + 23d] x 24 $$ = (½) x [2a + 23d] x 24 $$ = (½) x 75 x 24 $$ = 900

Looking on the sum of a1,a5,a10,a15,a20 and a24:

Sn = (n/2)(a1+an)

Using Sn = 225 and n =6,

we will get (a1+an) = 75

Using (a1 + an ) to compute for the sum of the first 24 terms with n=24

Sn = (n/2)(a1 + an)

Sn = (24/2)(75)

Sn = 900

an = a1 + (n-1)d & Sn = n/2 (2a1+(n-1)d) or n/2(a1+an) or (an+a1)(an-a1)/(a2-a1) for n=24 we find d=1 so S = 12(a24+a1) now a1+a5+a10+a15+a20+a24 = 225 or a1+a1+4+a1+9+a1+14+a1+19+a1+23=225 a1 = 26 & a24 = 49 S24 = 12(26+49) =900

Hello,

Given that 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

2a = 75 - 23d (1),

for sum of the 1st 24 terms,

S(24) = 12(2a + 23d),take (1) substitute into this S(24),

       = 12(75 -23d + 23d)

       =12(75)=900

therefore S(24) = 900,

thanks....