Question

17.The value of (13+23+33+...+153)- (1+2 +3 + ... +15) is​

Answer 1

GIVEN:

★(13+23+33+.... +153) -(1+2+3+4+.... +15)

TO FIND:

★The value of the given terms.

CONCEPT USED:

★We would simplify firstly the terms in the brackets separately.

★These are in Arthmetic Progression, so we would be using some AP related formulas.

ANSWER;

Let us take (13+23+33+... +153) as the first part &(1+2+3+4... +15) as second part for easy solving.

First part:

(13+23+33+43... +153)

We can see here that it is is in AP and the common difference is (23-13) =10

So, sum of n terms of AP is,

$\large\green{\boxed{S_{n}=\dfrac{n}{2}[2a+(n-1) d]}}$

Where • n is the number of terms in the AP • d is the common difference

• a is the first term

=>$S_{n}=\dfrac{15}{2}[2×13+(15-1) 10]$

=>$S_{n}=\dfrac{15}{2}[26+14×10]$

=>$S_{n}=\dfrac{15}{2}[26+140]$

=>$S_{n}=\dfrac{15}{\cancel{2}}×\cancel{166}$

. °. $S_{n}=1245$

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Now, in 2nd part

=(1+2+3+4+........ +15)

We can see that it is the sum of first n natural numbers.

So, sum if first n natural numbers is given by,

$\large\purple{\boxed{S_{natural}=n\dfrac{(n+1)}{2}}}$

Where • $S_{natural}$ is the sum of first n natural numbers.

=>$S_{natural}=n\dfrac{(n+1)}{2}$

=>$S_{natural}=15\dfrac{(15+1)}{2}$

=>$S_{natural}=15×\dfrac{\cancel{16}}{\cancel{2}}$

=>$S_{natural}=15\times8$

. °.$S_{natural}=120$

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Hence required answer

=$S_{n}-S_{natural}$

=$1245-120$

=$1125$

Hence the required answer is 1125.

$\huge\orange{\boxed{Answer:1125}}$

Answer 2

Answer :

675

Step-by-step explanation:

13+ 10n- 10 =1543

10n = 150

n = 15

12 + 10n - 10= 102

10n= 100

n= 10