Question

The sum of three terms is
of AP is 45.If 2nd term is 15,find the three terms

Answer 1

ᴄᴏɴᴄᴇᴘᴛ ᴜsᴇᴅ :-

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

Sᴏʟᴜᴛɪᴏɴ :-

Given That :-

→ S(3) = 45

→ (3/2)[2a + (3 - 1)d] = 45

→ 3[2a + 2d] = 45 * 2

→ 3*2*(a + d) = 45 * 2

→ 3(a + d) = 45

→ (a + d) = 15.

Also, Given That :-

→ T(2) = 15

→ a + (2 - 1)d = 15

→ (a + d) = 15 = same .

Now, Let the first three terms be (a-d), a, and (a+d).

And, we have given that, the sum of these terms is equal to 45.

So,

→ (a - d) + a + (a + d) = 45

→ 3a = 45

→ a = 15.

Putting this value we get,

→ (a + d) = 15

→ 15 + d = 15

→ d = 15 - 15

→ d = 0 .

Hence, we can conclude That, all Three - Terms of AP are 15 in this given case. (Or, Question is wrongly written.)

Answer 2

×× Mate your question is wrong ××

Please check it and ask it again

Let the 3 consecutive term = (a-d),a,(a+d)

$\tt s(3)=45\\\tt \frac{3}{2}[2a+(3-1)d=45\\\tt 3[2a+2d]=45×2\\\tt 3×2[a+d]】45×2\\\tt 6(a+d)=90\\\tt a+d=\frac{90}{6}\\\tt a+d=15$

And,

$\tt T(2)=15\\\tt a+(2-1)d=15\\a+d=15$

NOW,

$\tt (a-d)×a×(a+d)=45\\\tt 3a=45\\\tt a=\frac{45}{3}\\\tt a=15$

Putting a=15 in 3rd considered number

$\tt(a+d)=15\\\tt 15+d=15\\\tt d=15+15=0$