$\left (1 + 3 + 5 + \ldots + p^\text{th} \right ) \\ + \\ \left (1 + 3 + 5 + \ldots + q^\text{th} \right ) \\ = \\ \left (1 + 3 + 5 + \ldots + r^\text{th} \right )$

Given the above equation for positive integers $p,q,r$ with $p^\text{th}$ term greater than 39.

What is the smallest possible value of the expression below?

$\large p^\text{th} \ \text{term } + q^\text{th} \ \text{term } + r^\text{th} \ \text{term }$

The answer is 109.

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but pth term is 39 . hence 1+2p-2=39 p=20 satisfying triplet=20, 15and 25 hence answer must be =117 @jaiveer shekhawat

Aakash Khandelwal
- 6 years ago

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Yeah!! Exactly .. Was trying so hard to put 117 but it said it was wrong.... @Aakash you are right...

Jayesh Swami
- 5 years, 7 months ago

Actually I think the question pth term of the AP ishould be greater than 49.

Department 8
- 5 years, 5 months ago

can you elaborate why $24^{2} + 7^{2} = 25^{2}$ is the smallest triplet satisfying the condition. @jaiveer shekhawat

U Z
- 6 years, 7 months ago

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it's the smallest required pythagorean triplet because we are given that $p^{th}$ term ≥ 39, thus one of the number of the two left handed side number must be ≥ 39...

have a look at the possible triplets

now, you will observe that if

$p^{2}$ + $q^{2}$ = $r^{2}$

then

$p^{th}$ term = 2p-1

thus none of the above triplets are yielding us the result what we are getting from

$24^{2}$ + $7^{2}$ = $25^{2}$

thus , this is the least required triplet, if you consider triplets above this then you will find that

pth term + qth term +rth term will be maximised!!

jaiveer shekhawat
- 6 years, 7 months ago

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The question I see says " $p^\text{th}$ term equals to 39 "

NO GREATER THAN SIGN

The answer to what is on my computer screen is 117

Bob Kadylo
- 5 years, 5 months ago

If the pth term, qth term, and rth term are all greater than 39, q cannot equal 7.

Phi Li
- 5 years, 5 months ago

For p>39, the pythogorean triplet 40,9,41 is the smallest, 40^2+9^2=41^2 Hence smallest possible is 40+9+41=90 is the answer

Raja Srinivas
- 3 years, 9 months ago

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Sorry, 40 th p term is 79, llly q is 17 & r is 81

So answer should be 79+17+81=177

Raja Srinivas
- 3 years, 9 months ago

Given pth term is > 39, not p>39

Skanda Prasad
- 3 years, 6 months ago

jaiveer is right

Devendra Ram
- 3 years ago

Bro ans must be 90 (40^2+9^2=41^2) this is the smallest one among all triplets and true for given condition

Manohar Reddy
- 2 years, 12 months ago

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What we want to minimise is (pth term + qth term + rth term), not (p+q+r). The only restriction is pth term greater than 39, hence p>20, while q can be any positive integer. @jaiveer shekhawat has the correct answer: p=24, q=7, r=25 thus, (pth term + qth term + rth term)=47+13+49=109.

Thanos Petropoulos
- 2 years, 7 months ago

This seems wrong! According to question, $p>39$

Akihito Narihisago
- 1 year, 2 months ago

The trick is that the question says the pth term not the p itself thus jaiveer is right

Ahmed Sami
- 2 months, 2 weeks ago

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(1+3+5...p th term) = [ $\frac{p}{2}$ ].(2+(p-1)2)

= $p^{2}$

(1+3+5...q th term) = [ $\frac{q}{2}$ ].(2+(q-1)2)

= $q^{2}$

(1+3+5...r th term) = [ $\frac{r}{2}$ ].(2+(r-1)2)

= $r^{2}$

You will observe that it contains a Pythagorean triplet:

$p^{2}$ + $q^{2}$ = $r^{2}$

Thus, the smallest triplet satisfying our condition is :

$24^{2}$ + $7^{2}$ = $25^{2}$

Thus, 24 th term of first series is 47, the 7 th term of the second series is 13 and the 25 th term of the last series is 49.

Thus, the answer is 47+13+49= $\boxed{109}$