CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Question 24 An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three terms is 429. Find the AP. Given Number of terms of AP = 37 So, last term will be 37th term Middle most term = ((37 + 1))/2 = 38/2 = 19th term Theory Let AP be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 Here number of terms = 11 Middle term = 6 So, Middle term = ((11 + 1))/2 = 12/2 = 6 So, 19th term is middle most term Thus, 3 middle most terms will be 18th term, 19th term, 20th term 18th term = a18 = a + (18 – 1)d = a + 17d 19th term = a19 = a + (19 – 1)d = a + 18d 20th term = a20 = a + (20 – 1)d = a + 19d Given Sum of 3 middle most terms = 225 (a + 17d) + (a + 18d) + (a + 19d) = 225 (a + a + a) + (17d + 18d + 19d) = 225 3a + 54d = 225 3(a + 18d) = 225 a + 18d = 225/3 a + 18d = 75 Also, Sum of last 3 terms = 429 Last term = an = a37 = a + (37 – 1)d = a + 36d Second Last term = an – 1 = a37 – 1 = a36 = a + (36 – 1)d = a + 35d Third Last term = an – 2 = a37 – 2 = a35 = a + (35 – 1)d = a + 34d Now Sum of last 3 terms = 429 Putting values (a + 36d) + (a + 35d) + (a + 34d) = 429 (a + a + a) + (36d + 35d + 34d) = 429 3a + 105d = 429 3(a + 35d) = 429 a + 35d = 429/3 a + 35d = 143 Thus, our equations are a + 18d = 7 5 …(1) a + 35d = 143 …(2) Doing (2) – (1) (a + 35d) – (a + 18d) = 143 – 75 a – a + 35d – 18d = 143 – 75 17d = 68 d = 68/17 d = 4 From (1) a + 18d = 7 5 Putting d = 4 a + 18 × 4 = 75 a + 72 = 75 a = 75 – 72 a = 3 Thus, a = 3, d = 4 So, AP is 3, 7, 11, 15, 19, …… 147

6Thank You

ANSWER

Which term of the AP 24 ,21 ,18,15.....is the first negative twrm ?

Posted by Saumya Singh 7 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Gurman Singh 7 years, 9 months ago

Ops !it is 10the term and 10th term is-3

0Thank You

Sankalp Kumar 7 years, 9 months ago

10 th term

2Thank You

Gurman Singh 7 years, 9 months ago

-3

0Thank You

ANSWER

How do we find average of 2 given average

Posted by Harshit Chaudhry 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Kisi ko exam pata h

Posted by Garima Bajaj 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Teerth Raj Chaudhary 7 years, 9 months ago

Kiska

0Thank You

ANSWER

Cos2 67-sin2 23 what is the value

Posted by Rohan Dhall 7 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Teerth Raj Chaudhary 7 years, 9 months ago

Cos2(90 -23)-sin2 23 Sin2 23- sin2 23=0 Using identity cos(90-any angle)=sin any angle

1Thank You

Nyayir Riba 7 years, 9 months ago

It will be 0

2Thank You

Sakshi Sawant 7 years, 9 months ago

1Thank You

ANSWER

Given triangleABC _ triangle pqr if ab-pq=1|3 then find ar ∆abc|ar∆pqr

Posted by Suman Kumar 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Sakshi Sawant 7 years, 9 months ago

1/9

2Thank You

ANSWER

Preparation ho gai ky sbki?????

Posted by Tanisha Badlani 7 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Tanisha Badlani 7 years, 9 months ago

Or tumhari

0Thank You

Tanisha Badlani 7 years, 9 months ago

Hmm meri toh almost ho gai h...

0Thank You

Teerth Raj Chaudhary 7 years, 9 months ago

Tumhari ho gai paper kaise gaye tumhare

1Thank You

ANSWER

Are optional exercises no need to solve? Because they have mentioned: {NOT FROM EXAMINATION POINT OF VIEW}

Posted by Lucky Charan 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

(b-c)X2 +(c-a)X +(a-b)=0

Posted by Medhavi Anand 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Tan2a=cot (a- 18) Cot(90-2a)=cot(a-18) 108=3a a=36

Posted by Gurman Singh 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

If x=3 is one root of the quadratic equation x square - 2 kx-6= 0,then find value of k

Posted by Sanjay Vatsyayan 7 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Gurman Singh 7 years, 9 months ago

K=1/2

1Thank You

ANSWER

8.4ex

Posted by Adarsh Tripathi 7 years, 9 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Express the hcf of 963 and 657 as 963x and 65 7y where x and y are integers

Posted by Chetan More 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

By applying Euclid’s division lemma,
{tex}963 = 657\times 1 + 306.{/tex}
Since remainder ≠ 0, apply division lemma on divisor 657 and remainder 306
{tex}657 = 306\times 2 + 45.{/tex}
Since remainder ≠ 0, apply division lemma on divisor 306 and remainder 45
{tex}306 = 45\times 6 + 36.{/tex}
Since remainder ≠ 0, apply division lemma on divisor 45 and remainder 36
{tex}45 = 36\times 1 + 9.{/tex}
Since remainder ≠ 0, apply division lemma on divisor 36 and remainder 9
{tex}36 = 9 \times 4 + 0.{/tex}
Therefore, {tex}H.C.F. = 9.{/tex}

HCF of 2 numbers can be expressed as the linear combination of the numbers
{tex}\Rightarrow {/tex} 9 = 45 - 36 {tex}\times{/tex} 1
= {tex}45 - [306 - 45 \times 6] \times 1 = 45 - 306\times 1 + 45\times 6{/tex}
= {tex}45 \times 7 - 306\times 1 = [657 -306\times 2]\times 7 - 306 × 1{/tex}
= {tex}657 \times 7 - 306 \times 14 - 306 × 1{/tex}
= {tex}657\times 7 - 306\times 15{/tex}
= {tex}657\times 7 - [963 - 657 × 1] \times 15{/tex}
= 657 {tex}\times{/tex} 7 - 963 {tex}\times{/tex} 15 + 657 {tex}\times{/tex} 15
= 657 {tex}\times{/tex} 22 - 963 {tex}\times{/tex} 15.
Hence, obtained.