Question

solve the equation:-

$1 + 4 + 7 + ... + x = 925$
please don't scam for 5 points.​

Answer 1

Answer:

hey here is your solution

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being a science student I could also help you in commerce as I have a keen interest in maths especially in geometry and statistics

Step-by-step explanation:

$to \: find = \\ value \: of \: x \\ \\ so \: here \: given \: sequence \: is \\ 1,4,7, \: till \: x \\ \\ so \: here \: \\ t1 = 1 \\ t2 = 4 \\ t 3 = 7 \\ \\ hence \: then \\ d = t2 - t1 \\ = 4 - 1 \\ = 3 \\ \\ d = t3 - t2 \\ = 7 - 4 \\ = 3 \\ \\ hence \: as \: common \: difference \: is \: constant \\ the \: given \: sequence \: is \: an \: A.P \\ so \: for \: given \: A.P \\ a = t1 = 1 \\ d = 3 \\ Sn = 925 \\ tn = x \\ \\ so \: we \: know \: that \\ tn = a + (n - 1)d \\ \\ x = 1 + (n - 1) \times 3 \\ = 1 + 3n - 3 \\ 3n - 2 \\ x + 2 = 3n \\ \\ n = \frac{x + 2}{3} \\ \\ hence \: then \: using \\ Sn = \frac{n}{2} (t1 + tn) \\ \\ 925 \times 2 = \frac{x + 2}{3} \: (x + 1) \\ \\ 1850 \times 3 = (x + 2)(x + 1) \\ \\ 5550 = x {}^{2} + x + 2x + 2 \\ \\ x {}^{2} + 3x - 5548 = 0 \\ \\ so \: comparing \: it \: with \\ \: ax {}^{2} + bx + c = 0 \\ we \: get \\ \\ a = 1 \: , \: b = 3 \: ,c = - 5548$

$so \: now \: using \\ ∆ = b {}^{2} - 4ac \\ = (3) {}^{2} - 4 \times 1 \times ( - 5548) \\ = 9 - ( - 22192) \\ = 22192 + 9 \\ = 22201 \\ \\ so \: by \: applying \: formula \: method \\ we \: get \\ \\ x = \frac{ - b \: ± \: \sqrt{b {}^{2} - 4ac} }{2a} \\ \\ = \frac{ - 3 \: ± \: \sqrt{22201} }{(2 \times 1)} \\ \\ = \frac{ - 3 \:± \: 149 }{2} \\ \\ x = \frac{ - 3 \: + 149 }{2} \: \: or \: \: x = \frac{ - 3 - 149}{2} \\ \\ = \frac{146}{3} \: \: or \: \: = \frac{ - 152}{2} \\ \\ x= 73 \: \: or \: \: x = - 76$

$but \: if \: we \: consider \\ x = - 76 \\ \\ then \: \\ n = \frac{ - 76 + 2}{3} \\ \\ = \frac{ - 74}{3} \\ \\ so \: as \: total \: no \: of \: terms \: are \: never \: negative \\ as \: n \: ∈ \: N \\ x = - 76 \: \: is \: absurd$

$hence \: then \\ x = 73$

Answer 2

Answer:

• Value of x is 73.

Step-by-step explanation:

Given :

$\bold{1+4+7+..+x=925}$

To find :

Value of $\bold{x}$

Solution :

We see that the above series is an A.P

We have:

• First term (a) = 1

• Common difference (d) = 3

• Sₙ = 925

• Last term (l) = x

We know:

$\bold{S_n=\dfrac{n}{2}(2a+(n-1)d) }$

$\rightarrow \bold{925=\dfrac{n}{2}(2(1)+(n-1)3) }$

$\rightarrow \bold{925=\dfrac{n}{2}(2)+(3n-3) }$

$\rightarrow \bold{925=\dfrac{n}{2}(3n-1) }$

$\rightarrow \bold{1850=3n^2-n}$

Factorizing we get:

$\rightarrow \bold{3n^2-n-1850=0}$

$\rightarrow \bold{3n^2-75n+74n-1850=0}$

$\rightarrow \bold{3n(n-25)+74(n-25)=0}$

$\rightarrow \bold{3n+74=0\;and\;n-25=0}$

$\rightarrow \bold{n=-\dfrac{74}{3}\;and\;n=25 }$

∴ Value of $\bold{n=25}$

Now, $\bold{a+(n-1)d=x}$:

$\rightarrow \bold{1+(25-1)3=x}$

$\rightarrow \bold{1+24(3)=x}$

$\rightarrow \bold{1+72=x}$

∴ Value of $\bold{x=73}$