Question

Q.14a The sum of the first, third and seventeenth term of
term of an AP is 216.
Find the sum of the first 13 terms of the A.P.​

Answer 1

GIVEN:-

• The sum of the first, third and seventeenth term of an AP is 216.

SOLUTION:-

➡ let the first term of AP be 'a'.

➡ let the common difference of the AP be "d".

➡ T1 = a

➡ T3= a +2d

➡ T17 = a + 16d

➡ T1 +T3+T17 = 216

➡ 3a + 18d = 216 ----( both side divide by 3)

➡ .°. a + 6d = 72

➡ .°. T 7 = 72

NOW,

➡ sum of the first 13 term of AP.

➡ 13/2[2a + 12d]

➡ 6 × T 7

➡ 13 ×72

➡ .°. 936.

SO, the sum of the first 13 terms of the A.P.is 936.

Answer 2

$\bold{ Given:- \: \: 216} \begin{cases} \sf \tt{ \underline {Sum \: of \: all \: given \: terms}} \\ \sf{a_1 } \\ \sf{ a_3 } \\ \sf{a_{17} } \end{cases}$

$\\ \\ \tt \bold \red \star \: \underline{Solution:-}$

$\: \: \: \: \: \: \tt{a_1+a_3+a_{17}=216.}$

$\: \: \: \: \: \: \tt{a + a + 2d + a + 16d = 216. }$

$\: \: \: \: \: \: \tt{3a + 18d = 216.}$

$\tt{Divided \: by \: \red{ 3} \: both \: sides, we \: get}$

$\: \: \: \: \: \: \tt{ \frac{3a}{3} + \frac{18d}{3} = \frac{216}{3} .} \\ \\ \: \: \: \: \: \: \tt{ a + 6d = 72.} \\ \\ \: \: \: \: \: \: \tt{a_n = a + (n - 1)d. }\\ \\ \: \tt{ put \: n = 7.} \\ \\ \: \: \: \: \: \: \tt{72 = a + (7 - 1)d.} \\ \\ \: \: \: \: \: \: \tt{72 = a + 6d .}\\ \\ \: \: \: \: \: \: \tt{ a_7 = 72.}$

$\tt{Now, } \\ \: \: \: \: \: \: \tt{Sum \: of \: first \: 13 \: terms .}$

$\: \: \: \: \: \: \tt{S_n = \frac{n}{2} \{2a + (n - 1)d \}} \\ \\ \: \: \: \: \: \: \tt{ S_{13} = \frac{13}{2} \{2a + (13 - 1)d \}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{= \frac{13}{2} \{2a + 12d \}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \tt{ = \frac{13}{2} \times \frac{1}{2} \{ a + 6d \}} \\ \\\: \: \: \: \: \: \: \: \: \: \: \: \tt{= 13 \times 72. }\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \tt{= 936.}$