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 = $\frac{n}{2}(a+l)=\frac{8}{2}($