Question

1.} Find the common difference of an A.P. whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.

2.} If the sum of the first 7 terms of an A.P. is 119 and that of the first 17 terms is 714, find the sum of its first n terms.

Kindly don't spam!​

Answer 1

❶

GIVEN:

• First term (a) of an A.P = 5
• Sum of its first four terms is half the sum of the next four terms

TO FIND:

• What is the common difference of an AP ?

SOLUTION:

We have given that, the first term of an AP is 5

• a = 5

and,

Sum of first four terms is half the sum of the next four terms

❒ Sum of first four terms:

• 5, 5+d, 5+2d, 5+3d

❒  Sum of next four terms:

• 5+4d, 5+5d, 5+6d, 5+7d

According to question:-

$\sf{\longmapsto 5 + 5+d + 5+2d + 5+3d = \dfrac{1}{2} \times 5+4d + 5+5d + 5+6d + 5+7d }$

$\sf{\longmapsto 20 + 6d = \dfrac{1}{2} \times 20 + 22d }$

$\sf{\longmapsto 20 + 6d = \dfrac{1}{\cancel{2}} \times \cancel{2}(10+11d) }$

$\sf{\longmapsto 20 + 6d = 10+11d}$

$\sf{\longmapsto 20-10 = 11d-6d }$

$\sf{\longmapsto 10 = 5d }$

$\sf{\longmapsto d = \cancel\dfrac{10}{5}}$

$\large{\boxed{\bf{\star \: d = 2 \: \star}}}$

❝ Hence, the common difference is 2 ❞

______________________

❷

GIVEN:

• The sum of the first 7 terms of an A.P. is 119.
• The sum of the first 17 terms is 714

TO FIND:

• What is the sum of n terms ?

SOLUTION:

[CASE:- 1]

❍ The sum of the first 7 terms of an A.P. is 119.

We know that the formula for finding the sum of n terms is:-

$\large{\boxed{\bf{\star \: S_n = \dfrac{n}{2} [2a +(n-1) d] \: \star}}}$

$\sf{\longmapsto 119 = \dfrac{7}{2} [2 \times a + (7-1) d] }$

$\sf{\longmapsto 119 \times 2 = 7[2a + 6d ]}$

$\sf{\longmapsto 238 = 7[2a + 6 d] }$

$\bf{\longmapsto 238 = 14a + 42d...1)}$

$\sf{\longmapsto \dfrac{238 -42d}{14} = a }$

[CASE:- 2)]

❍ The sum of the first 17 terms is 714.

$\sf{\longmapsto 714 = \dfrac{17}{2} [2 \times a + (17-1) d] }$

$\sf{\longmapsto 714 \times 2 = 17[2a + 16d ]}$

$\sf{\longmapsto 1428 = 17[2a + 16 d] }$

$\bf{\longmapsto 1428 = 34a + 272d...2)}$

Put the value of a in equation 2)

$\sf{\longmapsto 1428 = 34 \Bigg( \dfrac{238 -42d}{14} \Bigg) + 272d }$

$\sf{\longmapsto 1428 = \dfrac{8092 - 1428d}{14} + 272d }$

$\sf{\longmapsto 1428 = \dfrac{8092 - 1428d + 3808d}{14} }$

$\sf{\longmapsto 1428 \times 14 = 8092+ 2380d }$

$\sf{\longmapsto 19992 - 8092 = 2380d }$

$\sf{\longmapsto \cancel{11900} = \cancel{2380}d }$

$\bf{\longmapsto 5 = d}$

Put the value of d in equation 1)

$\sf{\longmapsto 238 = 14a + 42 \times 5 }$

$\sf{\longmapsto 238 = 14a + 210 }$

$\sf{\longmapsto 238 - 210 = 14a }$

$\sf{\longmapsto 28 = 14a }$

$\bf{\longmapsto 2 = a }$

Sum of 'n' terms is:-

$\sf{\longmapsto S_n = \dfrac{n}{2} [2 \times 2 + (n-1)5]}$

$\sf{\longmapsto S_n = \dfrac{n}{2} [4 + 5n-5]}$

$\sf{\longmapsto S_n = \dfrac{n}{2} [ 5n-1]}$

$\bf{\longmapsto S_n = \dfrac{5n^2 - n}{2}}$

❝ Hence, the sum of first 'n' terms is 5n² – n/2 ❞

______________________

Answer 2

$\huge\underbrace\mathfrak\red{answer}$

1•let d is common difference of AP

now first 4term is 5,5+d,5+2d,5+3d

and next 4term 5+4d,5+5d,5+6d,5+7d

now according to question

20+6d=(20+22d)/2

20+6d=10+11d

d=2

2• is in above attachment

$<b><marquee>$

__________ꫝꪮρꫀ ⅈt ꫝꫀꪶρꫀᦔ ꪊ ________