Question

35. An arithmetic progression consists of 37 terms. The sum of the first 3 terms of
it is 12 and the sum of its last 3 terms is 318, then find the first and last terms
of the progression
​

Answer 1

Answer:

An AP consists of 37 terms. The sum of three middle most of terms is 225 and sum of the last three is 429. Find the A.P.

Sol:- Let the first term be 'a' and common difference be 'd' respectively.

Since the A.P. contains 37 terms. So, the middle most term is (37+1)/2th term =19th term.

Thus, three middle most terms of A.P are 18th, 19th and 20th terms.

Given a+(18−1)d+a(19−1)d+a(20−1)d=225

⇒a+17d+a+18d+a+19d=225

⇒3a+54d=225

⇒3(a+18d)=225

⇒a+18d=75

⇒a=75−18d ……….(1)

According to given information

a+(35−1)d+a+(36−1)d+a+(37−1)d=429

⇒a+34d+a+35d+a+36d=429

⇒3a+105d=429

⇒3(a+35d)=429

⇒(75−18d+35d)=143

⇒17d=143−75=68

⇒17d=68

⇒d=4 ∴ Thus, the A.P. 3,7,11,15,....

Put d=4 in (1)

a=75−(8(4))

a=75−72

a=3.

hope this answer helped you

Answer 2

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Gɪᴠᴇɴ :

• The AP consist of 37 terms

• Sum of first three terms is 12

• Sum of last three terms is 318.

Tᴏ Fɪɴᴅ :

• The common difference (d) of the given sequence.

• The first term (a) of the given sequence.

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

• Let first term of AP is 'a'. and common difference is 'd'.

According to statement,

• Sum of first three terms is 12.

$\rm :\implies\: \boxed{ \pink{ \bf \: a_{1} + a_{2} + a_{3}\: = \tt \: 12}}$

$\rm :\implies\:a + a + d + a + 2d = 12$

$\rm :\implies\:3a + 3d = 12$

$\rm :\implies\:a + d = 4$

$\rm :\implies\: \boxed{{ \pink{a = 4 - d }}}- - - (i)$

According to statement,

• Sum of last three terms is 318.

$\rm :\implies\: \boxed{ \pink{ \bf \: a_{37} + a_{36} + a_{35} \: = \tt \: 318}}$

$\rm :\implies\:a + 36d + a + 35d + a + 34d = 318$

$\rm :\implies\:3a + 105d = 318$

$\rm :\implies\:a + 35d = 106$

$\rm :\implies\:4 - d + 35d = 106 \: \: \: \: \: \: \: \: \: \: (\because \: a \: = 4 - d)$

$\rm :\implies\:34d = 102$

$\rm :\implies\: \boxed{ \green{ \bf \: d = 3}}$

On substituting the value of 'd' in equation (1), we get

$\rm :\implies\: \boxed{ \pink{ \bf \: a\: = \tt \:4 - 3 = 1 }}$