Arithmetic Progression in Līlāvatī

An arithmetic progression or arithmetic sequence (AP) is a sequence of numbers such that the difference from any succeeding term to its preceding term remains constant throughout the sequence. The constant difference is called common difference of that arithmetic progression. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common difference of 2.

Verse 123

सैकपदघ्नपदार्धमथैकाद्यङ्कयुतिः किल सङ्कलिताख्या ।

सा द्वियुतेन पदेन विनिघ्नी स्यात् त्रिहृता खलु सङ्कलितैक्यम् ॥ १२३ ॥

$1+2+3+...+n={\frac {n(n+1)}{2}}$

$1+(1+2)+(1+2+3)+...+(1+2+...+n)={\frac {n(n+1)(n+2)}{6}}$

Example

एकादीनां नवान्तानां पृथक् सङ्कलितानि मे ।

तेषां सङ्कलितैक्यानि प्रचक्ष्व गणक द्रुतम् ॥

Find $1+2+...+9$ and $1+(1+2)+...+(1+2+...+9)$

As per the verse 123 the answers are

$1+2+...+9={\frac {9(9+1)}{2}}={\frac {9X10}{2}}=45$

$1+(1+2)+...+(1+2+...+9)={\frac {9\ X\ (9+1)\ X\ (9+2)}{6}}={\frac {9\ X\ 10\ X\ 11}{6}}=165$

Verse 125

द्विघ्नपदं कुयुतं त्रिविभक्तं सङ्कलितेन हतं कृतियोगः ।

सङ्कलितस्य कृतेः सममेकाद्यङ्कघनैक्यमुदाहृतमाद्यैः॥

$1^{2}+2^{2}+......+n^{2}={\frac {n(n+1)(2n+1)}{6}}$

$1^{3}+2^{3}+......+n^{3}=\left[{\frac {n(n+1)}{2}}\right]^{2}$

Arithmetic Progression

व्येकपदघ्नचयो मुखयुक स्यादन्त्यधनम् मुखयुग्दलितं तम् ।

मध्यधनं पदसंगुणितं तत्सर्वधनं गणितं च तदुक्तम् ॥

If the first term is a and the common difference (C.D.) is d,

the n^th term of the Arithmetic Progression (A.P.) is given by $l=a+(n-1)d$

The sum of the n terms is given by ${\frac {n}{2}}(a+l)$. Here ${\frac {a+l}{2}}$ is the middle term.

Example 1

तेषामेव च वर्गैक्यं घनैक्यं च वद द्रुतम् ।

कृतिसङ्कलनामार्गे कुशला यदि ते मतिः ॥

Tell me the sums of 1²+...+9² and 1³ + . . . +9³.

Comment:

$1^{2}+......+9^{2}={\frac {9(9+1)(2\ X\ 9+1)}{6}}={\frac {9(10)(19)}{6}}=285$

$1^{3}+......+9^{3}=\left[{\frac {9(9+1)}{2}}\right]^{2}=\left[{\frac {9(10)}{2}}\right]^{2}=2025$

Example 2

आदिः सप्त चयः पंच गच्छोऽष्टौ यत्र तत्र मे ।

मध्यान्त्यधन-संख्ये के वद सर्वधनं च किम् ॥

If the first term of an A.P. is 7 the common difference is 5 and the number of terms is 8, find the middle term, the last term and the sum.

Comment:

first term a = 7 ; common difference d = 5 : number of terms n = 8

last term $l=a+(n-1)d=7+(8-1)5=7+(7)5=42$

middle term = ${\frac {a+l}{2}}={\frac {7+42}{2}}={\frac {49}{2}}$ which is not a term in the A.P.

Sum = ${\frac {n}{2}}(a+l)={\frac {8}{2}}(7+42)=4\ X\ 49=196$

Example 3

आद्ये दिने द्रम्मचतुष्टयं यो दत्त्वा दिनेभ्योऽनुदिनं प्रवृत्तः ।

दातुं सखे पंचचयेन पक्षे द्रम्मा वद द्राक्कति तेन दत्ताः ॥

A gentleman gave 4 D (drammas) as charity (to a Brahmin) on the first day. For fifteen days he continued his charity, everyday increasing his contribution by 5 D over the previous day. Find out the total charity.

Comment: Here a = 4, common difference = d = 5, n = 15.

Last term $l=a+(n-1)d=4+(15-1)5=4+(14)5=74$

Middle term $m={\frac {a+l}{2}}={\frac {4+74}{2}}={\frac {78}{2}}=39$

Total charity = ${\frac {15}{2}}(4+74)={\frac {15\ X\ 78}{2}}=15\ X\ 39=585$ D

Formula to find the first term of an Arithmetic Progression

गच्छहृते गणिते वदनं स्यात् व्येकपदघ्नचयार्धविहीने ॥

To find the first term of an A.P., divide the given sum by the number of terms. Multiply the number of terms minus one by half the common difference, and subtract this result from the quotient already obtained.

$a={\frac {S}{n}}-{\frac {(n-1)d}{2}}$

Example

पञ्चाधिकं शतं श्रेढीफलं सप्तपदं किल ।

चयं त्रयं वयं विद्मोवदनं वद नंदन ॥

O son, find the mouth (the first term a) of an A.P whose S = 105, n = 7 , d =3

Comment: $a={\frac {S}{n}}-{\frac {(n-1)d}{2}}$ = ${\frac {105}{7}}-{\frac {(7-1)\ X\ 3}{2}}=15-{\frac {6\ X\ 3}{2}}=15-9=6$

Formula to find common difference (C.D) in an Arithmetic Progression

गच्छहृतं धनमादिविहीनं व्येकपदार्धहृतं च चयः स्यात् ॥

Divide the sum by the number of terms (of an A.P.) and subtract the first term from this quotient. This result divided by half of the number

of terms minus one is the common difference.

$d={\frac {{\frac {S}{n}}-a}{\frac {n-1}{2}}}$

Example

प्रथममगमदह्ना योजने यो जनेश:

तदनु ननु कयाऽसौ ब्रूहि यातोऽध्ववृद्ध्या ।

अरिकरिहरणार्थं योजनानामशीत्या

रिपुनगरमवाप्तः सप्तरात्रेण धीमन् ॥

To capture enemy elephants, a king covers 2 Y (yojanas) on the first day and then increases his distance by A.P. on subsequent days. If he travels 80 Y in 7 days, O intelligent boy, find out the extra distance each day.

Comment: S = 80 , n=7 , a = 2

$d={\frac {{\frac {S}{n}}-a}{\frac {n-1}{2}}}$$={\frac {{\frac {80}{7}}-2}{\frac {7-1}{2}}}={\frac {66}{7}}\ X\ {\frac {2}{6}}={\frac {22}{7}}=3{\frac {1}{7}}$Y

Formula to find the number of terms of an A.P.

श्रेढीफलादुत्तरलोचनघ्नात् चर्यार्धवक्त्रान्तरवर्गयुक्तात् ।

मूलं मुखोनं चयखंडयुक्तम् चयोद्धृतं गच्छमुदाहरन्ति ॥

Add the square of the difference between the first term and half the C.D. to the sum (of the given A.P.) multiplied by twice the C.D. Then find (the positive) square root of this result. Subtract the first term from and add half the C.D. to this square root. The resultant divided by the C.D. is number of terms (in the A.P.).

$n={\frac {{\sqrt {2Sd+(a-{\frac {d}{2}})^{2}}}-(a-{\frac {d}{2}})}{d}}$

Example

द्रम्मत्रयं यः प्रथमेऽह्नि दत्त्वा दातुं प्रवृत्तो द्विचयेन तेन ।

शतत्रयं षष्ट्यधिकं द्विजेभ्यो दत्तं कियद्भिर्दिवसैर्वदाशु ॥

A donor gave 3 D (drammas) in charity to a Brahmin on the first day. He continued increasing his donation each day by 2 D. If the total amount paid by him equals 360 D, how many days did he give in charity?

Comment:

$n={\frac {{\sqrt {2Sd+(a-{\frac {d}{2}})^{2}}}-(a-{\frac {d}{2}})}{d}}$

$n={\frac {{\sqrt {2\ X\ 360\ X\ 2+(3-{\frac {2}{2}})^{2}}}-(3-{\frac {2}{2}})}{2}}$

$n={\frac {{\sqrt {1440+4}}-2}{2}}={\frac {38-2}{2}}=18$

See Also

लीलावती में 'समांतर श्रेढ़ी'